Let's work through the problem step by step: ### 1. Determine Reactor Inlet Concentration Let \( C_{A} \) be the concentration in the fresh feed, \( C_{AR} \) be the concentration at the reactor inlet, and \( C_A \) be the reactor exit concentration. Recycle ratio \( R = \frac{F_R}{F_F} = 1.5 \). Total flow to reactor: \( F = F_F + F_R = F_F(1 + R) = .5 \times (1 + 1.5) = 1.25 \) m³/min. The inlet concentration to the reactor is: \[ C_{AR} = \frac{F_F C_{A} + F_R C_A}{F} \] \[ C_{AR} = \frac{.5 \times 4 + (.5 \times 1.5) \times C_A}{1.25} \] \[ C_{AR} = \frac{2 + .75 C_A}{1.25} \] ### 2. Write CSTR Design Equation At steady state for a first-order reaction: \[ V = \frac{F(C_{AR} - C_A)}{k C_A} \] Given \( k = .8 \) min\(^{-1} \), \( V = 1.5 \) m³, and \( F = 1.25 \) m³/min. So, \[ 1.5 = \frac{1.25 (C_{AR} - C_A)}{.8 C_A} \] \[ 1.5 \times .8 C_A = 1.25 C_{AR} - 1.25 C_A \] \[ 1.2 C_A + 1.25 C_A = 1.25 C_{AR} \] \[ 2.45 C_A = 1.25 C_{AR} \] \[ C_A = \frac{1.25}{2.45} C_{AR} \] \[ C_A = .5102 C_{AR} \] ### 3. Find Exit Concentration \( C_A \) Recall from earlier: \( C_{AR} = \frac{2 + .75 C_A}{1.25} \). Now substitute the expression for \( C_A \) in terms of \( C_{AR} \): Let \( C_A = .5102 C_{AR} \): \[ C_{AR} = \frac{2 + .75 \times .5102 C_{AR}}{1.25} \] \[ C_{AR} = \frac{2 + .38265 C_{AR}}{1.25} \] \[ 1.25 C_{AR} = 2 + .38265 C_{AR} \] \[ 1.25 C_{AR} - .38265 C_{AR} = 2 \] \[ .86735 C_{AR} = 2 \] \[ C_{AR} = \frac{2}{.86735} = 2.305 \text{ kmol/m}^3 \] Now, \( C_A = .5102 \times 2.305 = 1.177 \text{ kmol/m}^3 \) ### 4. Overall Conversion Overall conversion: \[ X = \frac{C_{A} - C_A}{C_{A}} \] \[ X = \frac{4 - 1.177}{4} = \frac{2.823}{4} = .7058 \] So, overall conversion is **70.6%**. ### 5. Effect if Recycle Ratio Doubles If \( R \) doubles, \( R = 3. \): New total flow: \( F = .5 \times (1 + 3) = 2. \) m³/min. New expressions: \[ C_{AR} = \frac{.5 \times 4 + .5 \times 3 \times C_A}{2.} = \frac{2 + 1.5 C_A}{2.} \] CSTR design equation: \[ 1.5 = \frac{2. (C_{AR} - C_A)}{.8 C_A} \] \[ 1.5 \times .8 C_A = 2. C_{AR} - 2. C_A \] \[ 1.2 C_A + 2. C_A = 2. C_{AR} \] \[ 3.2 C_A = 2. C_{AR} \] \[ C_A = \frac{2.}{3.2} C_{AR} = .625 C_{AR} \] Now substitute: \[ C_{AR} = \frac{2 + 1.5 \times .625 C_{AR}}{2.} \] \[ C_{AR} = \frac{2 + .9375 C_{AR}}{2.} \] \[ 2 C_{AR} = 2 + .9375 C_{AR} \] \[ 2 C_{AR} - .9375 C_{AR} = 2 \] \[ 1.0625 C_{AR} = 2 \] \[ C_{AR} = \frac{2}{1.0625} = 1.882 \text{ kmol/m}^3 \] \[ C_A = .625 \times 1.882 = 1.176 \text{ kmol/m}^3 \] Overall conversion: \[ X = \frac{4 - 1.176}{4} = .706 \] Thus, doubling the recycle ratio barely changes the exit concentration or conversion (from about 70.6% to 70.6%), indicating diminishing returns for increasing recycle at these conditions. --- **Summary of answers:** 1. Reactor inlet concentration \( C_{AR} = 2.305 \) kmol/m³. 2. CSTR design equation: \( V = \frac{F(C_{AR} - C_A)}{k C_A} \). 3. Exit concentration \( C_A = 1.177 \) kmol/m³. 4. Overall conversion \( X = 70.6\% \). 5. If recycle ratio doubles: \( C_A = 1.176 \) kmol/m³, \( X = 70.6\% \) (almost unchanged).