Let's solve each part step by step: **1. CSTR Design Equation** For a steady-state, isothermal CSTR with a first-order reaction: \[ \text{Mole balance:} \quad F(C_{A} - C_A) = V r_A \] Given the rate law: \( r_A = -k C_A \) \[ F(C_{A} - C_A) = V (k C_A) \] \[ C_{A} - C_A = \frac{V}{F} k C_A \] \[ C_{A} = C_A \left(1 + \frac{V}{F}k\right) \] \[ C_A = \frac{C_{A}}{1 + \frac{V}{F}k} \] --- **2. Calculate Outlet Concentration of A** Given: \( C_{A} = 5 \) kmol/m³ \( V = 1.5 \) m³ \( F = .6 \) m³/min \( k = .9 \) min⁻¹ Substitute into the equation: \[ C_A = \frac{5}{1 + \frac{1.5}{.6} \times .9} \] First, calculate \(\frac{1.5}{.6} = 2.5\). \[ C_A = \frac{5}{1 + (2.5 \times .9)} \] \[ 2.5 \times .9 = 2.25 \] \[ C_A = \frac{5}{1 + 2.25} = \frac{5}{3.25} \approx 1.538\, \text{kmol/m}^3 \] --- **3. Calculate Conversion of A** Conversion, \( X \), is given by: \[ X = \frac{C_{A} - C_A}{C_{A}} \] \[ X = \frac{5 - 1.538}{5} = \frac{3.462}{5} = .692 \] So, the conversion is \( 69.2\% \). --- **4. Calculate Space Time** Space time, \( \tau \), is: \[ \tau = \frac{V}{F} \] \[ \tau = \frac{1.5}{.6} = 2.5\, \text{min} \] --- **5. Reactor Volume for 80% Conversion** For 80% conversion, \( X = .8 \). \[ C_A = C_{A} (1 - X) = 5 \times (1 - .8) = 1\, \text{kmol/m}^3 \] Substitute in the CSTR design equation: \[ C_A = \frac{C_{A}}{1 + \frac{V}{F}k} \] \[ 1 = \frac{5}{1 + \frac{V}{.6} \times .9} \] \[ 1 + \frac{V}{.6} \times .9 = 5 \] \[ \frac{V}{.6} \times .9 = 4 \] \[ \frac{V}{.6} = \frac{4}{.9} \] \[ V = .6 \times \frac{4}{.9} = .6 \times 4.444 = 2.667\, \text{m}^3 \] --- **Final Answers** 1. CSTR design equation: \( C_A = \frac{C_{A}}{1 + \frac{V}{F}k} \) 2. Outlet concentration of A: \( 1.54\, \text{kmol/m}^3 \) 3. Conversion of A: \( 69.2\% \) 4. Space time: \( 2.5\, \text{min} \) 5. Reactor volume for 80% conversion: \( 2.67\, \text{m}^3 \)