Let's solve the CSTR problem step-by-step: --- ## **Given Data** - **Volume of reactor, \( V \) = 1.5 m³** - **Inlet concentration of A, \( C_{A} = 2 \) mol/L = 200 mol/m³** (since 1 L = .001 m³) - **Volumetric flow rate, \( q = .1 \) m³/s** - **First-order rate constant, \( k = .5 \) s⁻¹** --- ## **Step 1: Material Balance for CSTR (Steady-State)** \[ \text{General mole balance: } \frac{dN_A}{dt} = F_{A} - F_A + r_A V \] At steady-state, \(\frac{dN_A}{dt} = \): \[ = F_{A} - F_A + r_A V \] Where: - \( F_{A} = \) inlet molar flow rate of A \( = q \cdot C_{A} \) - \( F_A = \) outlet molar flow rate of A \( = q \cdot C_A \) - \( r_A = \) rate of reaction of A (disappearance) \( = -k C_A \) - \( V = \) volume of reactor --- ## **Step 2: Substitute Values** \[ = q C_{A} - q C_A - k C_A V \] \[ q C_{A} = q C_A + k C_A V \] \[ q C_{A} = C_A (q + k V) \] \[ C_A = \frac{q C_{A}}{q + k V} \] --- ### **Plug in Numbers** - \( q = .1 \) m³/s - \( C_{A} = 200 \) mol/m³ - \( k = .5 \) s⁻¹ - \( V = 1.5 \) m³ \[ C_A = \frac{.1 \times 200}{.1 + .5 \times 1.5} \] \[ C_A = \frac{200}{.1 + .75} \] \[ C_A = \frac{200}{.85} \approx 235.29 \text{ mol/m}^3 \] --- ## **Step 3: Calculate Conversion of A** \[ X = \frac{C_{A} - C_A}{C_{A}} \] \[ X = \frac{200 - 235.29}{200} \] \[ X = \frac{1764.71}{200} = .882 \] So, **conversion of A is approximately 88.2%** --- ## **Step 4: Outlet Concentration of B ( \( C_B \) )** For every mole of A consumed, 1 mole of B is produced (from stoichiometry): \[ C_B = C_{A} - C_A \] \[ C_B = 200 - 235.29 = 1764.71 \text{ mol/m}^3 \] --- ## **Final Answers** ### **1. Steady-state conversion of A:** \[ \boxed{X = .882 \ \text{or} \ 88.2\%} \] ### **2. Outlet concentration of A:** \[ \boxed{C_A = 235.29 \ \text{mol/m}^3} \] ### **3. Outlet concentration of B:** \[ \boxed{C_B = 1764.71 \ \text{mol/m}^3} \] --- **Summary Table:** | Quantity | Value | |----------------------------|-------------------| | Conversion of A (\(X\)) | .882 (88.2%) | | Outlet \(C_A\) | 235.29 mol/m³ | | Outlet \(C_B\) | 1764.71 mol/m³ |