# Step-by-Step Solution Let's solve each task as per the problem statement. ## 1. Write Equilibrium Conversion Expression For the reaction \( A \leftrightarrow B \): At equilibrium, \( K = \frac{C_{B,eq}}{C_{A,eq}} \). If \( X_{eq} \) is the equilibrium conversion, then at equilibrium: - \( C_{A,eq} = C_{A}(1 - X_{eq}) \) - \( C_{B,eq} = C_{A}X_{eq} \) So, \[ K = \frac{C_{B,eq}}{C_{A,eq}} = \frac{C_{A}X_{eq}}{C_{A}(1 - X_{eq})} = \frac{X_{eq}}{1 - X_{eq}} \] Solving for \( X_{eq} \): \[ K = \frac{X_{eq}}{1 - X_{eq}} \implies K(1 - X_{eq}) = X_{eq} \implies K - KX_{eq} = X_{eq} \implies K = X_{eq}(1 + K) \implies X_{eq} = \frac{K}{1 + K} \] Given \( K = 4 \): \[ X_{eq} = \frac{4}{1 + 4} = \frac{4}{5} = .8 \] ## 2. Write CSTR Design Equation Including Equilibrium Limitation The CSTR mole balance for A: \[ F_{A} - F_A = r_AV \] Substitute \( F_{A} = C_{A}F \), \( F_A = C_A F \), and \( r_A = -r_f + r_r \). The forward rate is \( r_f = kC_A \), and the reverse rate is \( r_r = kC_B/K \) (from detailed balance). Net rate: \[ r_A = kC_A - k\frac{C_B}{K} \] But \( C_A = C_{A}(1-X) \) and \( C_B = C_{A}X \). So, \[ r_A = kC_{A}(1-X) - k\frac{C_{A}X}{K} \] \[ = kC_{A}[(1-X) - \frac{X}{K}] \] CSTR balance: \[ C_{A}F - C_A F = r_AV \] \[ C_{A}F - C_{A}(1-X)F = r_AV \] \[ C_{A} F X = r_A V \] \[ C_{A} F X = kC_{A}[(1-X) - \frac{X}{K}]V \] Divide both sides by \( C_{A} \): \[ F X = k[(1-X) - \frac{X}{K}]V \] \[ F X = kV\left[1 - X - \frac{X}{K}\right] \] \[ F X = kV\left[1 - X\left(1 + \frac{1}{K}\right)\right] \] ## 3. Calculate Actual Conversion in Reactor Plug in values: - \( F = .6\, \text{m}^3/\text{min} \) - \( k = .8\, \text{min}^{-1} \) - \( V = 1.5\, \text{m}^3 \) - \( K = 4 \) So, \[ F X = kV\left[1 - X\left(1 + \frac{1}{K}\right)\right] \] \[ .6 X = .8 \times 1.5 \left[1 - X\left(1 + \frac{1}{4}\right)\right] \] \[ .6 X = 1.2 \left[1 - X\left(1.25\right)\right] \] \[ .6 X = 1.2 - 1.2 \times 1.25 X \] \[ .6 X = 1.2 - 1.5 X \] \[ .6 X + 1.5 X = 1.2 \] \[ 2.1 X = 1.2 \] \[ X = \frac{1.2}{2.1} \approx .571 \] ## 4. Calculate Exit Temperature Adiabatic operation means the temperature change is due to the reaction. If the heat of reaction and heat capacities were provided, we could calculate this. However, the problem does **not** provide enthalpy data, so the exit temperature remains at the feed temperature of 300 K. ## 5. Compare Equilibrium Conversion vs Actual Conversion Equilibrium conversion: \( X_{eq} = .8 \). Actual conversion in CSTR: \( X_{CSTR} \approx .571 \). The actual conversion in the CSTR is significantly less than the equilibrium conversion due to finite reactor size and the nature of CSTR operation. ## 6. Effect of Increasing Reactor Volume Increasing the reactor volume \( V \) increases the term \( kV \) in the design equation, thus increasing the conversion \( X \). As \( V \) approaches infinity, the conversion \( X \) approaches the equilibrium conversion \( X_{eq} \). In practice, this means larger reactors yield higher conversions, but they can never exceed the equilibrium limit. --- ## **Summary Table** | Quantity | Value | |-------------------------------|------------| | Equilibrium conversion, \( X_{eq} \) | .8 | | Actual CSTR conversion, \( X \) | .571 | | Exit temperature, \( T_{exit} \) | 300 K | --- **Final Answers:** Equilibrium conversion \( X_{eq} = .8 \). CSTR actual conversion \( X_{CSTR} \approx .571 \). Exit temperature = 300 K (no heat effect data given). Increasing reactor volume increases conversion, approaching \( X_{eq} \) at very large volumes. CSTR conversion is always less than equilibrium conversion for finite volume.