Let's solve the problem step by step: --- ## **Given Data** - **Reaction:** \( A \rightarrow R + S \) - **Rate equation:** \(-r_A = \frac{k C_A^{1/2}}{1 + K_A C_A}\) - \( k = 10\, (\text{kmol/m}^3)^{.5} \text{h}^{-1} \) - \( K_A = 16\, \text{m}^3/\text{kmol} \) - \( F_{A} = 100\, \text{kmol/h} \) - \( C_{A} = .25\, \text{kmol/m}^3 \) - **Required conversion:** \( X = .90 \) --- ## **1. Reactor Configuration for Minimum Volume** ### **Step 1: Reactor Type Selection** For minimum volume at high conversion, **plug flow reactor (PFR)** is typically preferred over a single CSTR, because PFRs are more efficient for first-order and similar reactions. For this rate law, let's check this by comparison at high conversion: #### **Why?** - For reactions where the rate decreases with conversion (as here, since denominator increases), PFR is more volume-efficient. - A series of CSTRs approaches PFR behavior and reduces volume compared to a single CSTR. **Conclusion:** **PFR** is suggested for minimum volume. --- ## **2. Calculate Reactor Volume for 90% Conversion in a PFR** ### **Step 2: Write the Mole Balance for a PFR** \[ \text{For a PFR:} \quad F_{A} \int_{}^{X} \frac{dX}{-r_A} = V \] - \( F_{A} = 100\, \text{kmol/h} \) - \( X = .90 \) ### **Step 3: Express \(-r_A\) in Terms of \(X\)** \[ C_A = C_{A}(1 - X) \] \[ -r_A = \frac{k C_A^{1/2}}{1 + K_A C_A} \] \[ = \frac{k [C_{A}(1-X)]^{1/2}}{1 + K_A [C_{A}(1-X)]} \] ### **Step 4: Substitute and Set Up the Integral** \[ V = F_{A} \int_{}^{.9} \frac{dX}{\frac{k [C_{A}(1-X)]^{1/2}}{1 + K_A C_{A}(1-X)}} \] \[ = F_{A} \int_{}^{.9} \frac{1 + K_A C_{A}(1-X)}{k [C_{A}(1-X)]^{1/2}} dX \] #### **Plug in values:** - \( k = 10 \) - \( K_A = 16 \) - \( C_{A} = .25 \) - \( F_{A} = 100 \) \[ V = 100 \int_{}^{.9} \frac{1 + 16 \times .25 \times (1-X)}{10 \times [.25(1-X)]^{1/2}} dX \] \[ = 100 \int_{}^{.9} \frac{1 + 4(1-X)}{10 \times [.25(1-X)]^{1/2}} dX \] \[ = 100 \int_{}^{.9} \frac{1 + 4 - 4X}{10 \times (.25)^{.5}(1-X)^{.5}} dX \] \[ = 100 \int_{}^{.9} \frac{5 - 4X}{10 \times .5 \times (1-X)^{.5}} dX \] \[ = 100 \int_{}^{.9} \frac{5 - 4X}{5 (1-X)^{.5}} dX \] \[ = 100 \int_{}^{.9} \frac{5 - 4X}{5 (1-X)^{.5}} dX \] \[ = 20 \int_{}^{.9} \frac{5 - 4X}{5 (1-X)^{.5}} dX \] \[ = 4 \int_{}^{.9} \frac{5 - 4X}{5 (1-X)^{.5}} dX \] \[ = 4 \int_{}^{.9} \left( \frac{1}{(1-X)^{.5}} - \frac{4X}{5(1-X)^{.5}} \right) dX \] --- ### **Step 5: Solve the Integral** \[ V = 4 \int_{}^{.9} (1-X)^{-.5} dX - 4 \int_{}^{.9} \frac{4X}{5}(1-X)^{-.5} dX \] \[ = 4 \left[ \int_{}^{.9} (1-X)^{-.5} dX - \frac{4}{5} \int_{}^{.9} X(1-X)^{-.5} dX \right] \] #### **First Integral:** \[ \int (1-X)^{-.5} dX = -2(1-X)^{.5} \] So, \[ \int_{}^{.9} (1-X)^{-.5} dX = -2[(1-.9)^{.5} - (1-)^{.5}] = -2[(.1)^{.5} - 1] \] #### **Second Integral:** Let’s use integration by parts, let \( u = X \), \( dv = (1-X)^{-.5} dX \). - \( du = dX \) - \( v = -2(1-X)^{.5} \) So, \[ \int X(1-X)^{-.5} dX = X \cdot (-2(1-X)^{.5}) - \int (-2(1-X)^{.5}) dX \] \[ = -2X(1-X)^{.5} + 2\int (1-X)^{.5} dX \] \[ \int (1-X)^{.5} dX = -\frac{2}{3}(1-X)^{1.5} \] So, \[ = -2X(1-X)^{.5} + 2 \left[ -\frac{2}{3}(1-X)^{1.5} \right] = -2X(1-X)^{.5} - \frac{4}{3}(1-X)^{1.5} \] So, \[ \int_{}^{.9} X(1-X)^{-.5} dX = \left[ -2X(1-X)^{.5} - \frac{4}{3}(1-X)^{1.5} \right]_{}^{.9} \] --- ### **Step 6: Substitute Limits** #### **First Integral** \[ -2[(.1)^{.5} - 1] = -2[.3162 - 1] = -2[-.6838] = 1.3676 \] #### **Second Integral** Calculate at \( X = .9 \): - \( (1-X) = .1 \) - \( (.1)^{.5} = .3162 \) - \( (.1)^{1.5} = .1 \times .3162 = .03162 \) So at \( X = .9 \): \[ -2 \times .9 \times .3162 = -.5692 \] \[ -\frac{4}{3} \times .03162 = -.04216 \] Sum: \( -.5692 - .04216 = -.61136 \) At \( X = \): - \( (1-)^{.5} = 1 \) - \( (1-)^{1.5} = 1 \) \[ -2 \times \times 1 = \] \[ -\frac{4}{3} \times 1 = -1.3333 \] Sum: \( - 1.3333 = -1.3333 \) So the definite integral: \[ (-.61136) - (-1.3333) = .72194 \] Multiply by \( \frac{4}{5} = .8 \): \[ .8 \times .72194 = .57755 \] --- ### **Step 7: Final Calculation for Volume \( V \)** Combine results: \[ V = 4 \left[ 1.3676 - .57755 \right] = 4 \times .79005 = 3.1602 \ \text{m}^3 \] --- ## **Summary** ### **a) Reactor Configuration** - **A single Plug Flow Reactor (PFR)** is suggested for minimum volume. ### **b) Required Reactor Volume** - **Total reactor volume for 90% conversion:** \[ \boxed{3.16\ \text{m}^3} \] --- ### **Final Answers** **a)** Suggest PFR for minimum volume. **b)** Required reactor volume: \(\boxed{3.16\ \text{m}^3}\)