Let's provide a clear, step-by-step, fully explained solution for **Question 2** as requested: --- # 2. Analytical Expressions for Gas-Phase Polymerization in a Plug Flow Reactor ### a) **Derive the differential design equation for catalyst weight** For a plug flow reactor (PFR) with catalyst weight \( W \) as the independent variable and fractional conversion \( X \), the design equation is: \[ \frac{dX}{dW} = -r' \] Rearranging and strictly omitting any numerical integration, the total catalyst mass required for a given conversion is: \[ W = \int \frac{dX}{-r'} \] This is the fundamental symbolic relationship between catalyst weight and fractional conversion for a PFR. --- ### b) **Explicit algebraic relationship: Pressure ratio as a function of catalyst mass** Given the pressure ratio as a function of catalyst mass \( W \): \[ y = (1 - \alpha \cdot W)^{.5} \] Where: - \( y \) = pressure ratio (local pressure/initial pressure), - \( \alpha \) = proportionality constant, - \( W \) = catalyst mass. This algebraic expression shows how the pressure ratio falls as more catalyst mass is traversed. --- ### c) **Modified kinetic rate law including pressure-induced volumetric expansion** The modified rate law accounting for pressure change and volumetric expansion is: \[ -r' = k \cdot C_{A} \cdot \frac{1 - X}{1 + \epsilon X} \cdot y \] Where: - \( k \) = rate constant, - \( C_{A} \) = initial concentration of A, - \( X \) = fractional conversion, - \( \epsilon \) = fractional change in moles (expansion parameter), - \( y \) = pressure ratio (from part b). --- ### d) **Final integrated boundary condition: Ultimate maximum conversion** As internal absolute pressure approaches zero, \( y \to \). The theoretical maximum conversion \( X_{\text{max}} \) can be found by setting \( y = \): From part b: \[ y = (1 - \alpha W)^{.5} \] For \( y \to \), \( 1 - \alpha W \to \implies W_{\text{max}} = \frac{1}{\alpha} \) Substitute \( y \to \) into the rate law from part c: \[ -r' \to \] At this point, the reaction ceases since the driving force (pressure) is zero. The ultimate theoretical maximum conversion is thus reached just as \( y \to \), or equivalently, \( W \to \frac{1}{\alpha} \). --- ## **Final Answers Summary** **a)** \( W = \int \frac{dX}{-r'} \) **b)** \( y = (1 - \alpha W)^{.5} \) **c)** \( -r' = k \cdot C_{A} \cdot \frac{1 - X}{1 + \epsilon X} \cdot y \) **d)** The ultimate maximum conversion is achieved as \( W \to \frac{1}{\alpha} \), when \( y \to \), and the rate drops to zero. --- If you need the solution for question 3, let me know!

# 3. Non-Ideal Tubular Chemical Reactor Analysis ### Step 1: Evaluate the Dimensionless Vessel Dispersion Number To characterize the longitudinal mixing intensity in a non-ideal tubular reactor, we calculate the dimensionless dispersion number \( D^* \): \[ D^* = \frac{D}{L} \cdot \frac{1}{\tau} \] Where: - \( D \) = axial dispersion coefficient, - \( L \) = length of the reactor, - \( \tau \) = mean residence time. Given that the mean residence time \( \tau = 45 \) seconds, we need to evaluate \( D \) and \( L \). In cases where specific values for \( D \) and \( L \) are not provided, we assume standard conditions to illustrate the calculation. Assuming \( D = 0.1 \, \text{m}^2/\text{s} \) and \( L = 1 \, \text{m} \) as an example: \[ D^* = \frac{0.1}{1} \cdot \frac{1}{45} = \frac{0.1}{45} \approx 0.00222 \] ### Step 2: Calculate the Corrected Overall Chemical Conversion For a first-order reaction, the corrected overall conversion \( X \) can be expressed using the relationship: \[ X = 1 - e^{-k \tau} \] Given the reaction rate constant \( k = 0.03 \, \text{s}^{-1} \) and \( \tau = 45 \, \text{s} \): \[ X = 1 - e^{-0.03 \times 45} \] Calculating the exponent: \[ 0.03 \times 45 = 1.35 \] Thus: \[ X = 1 - e^{-1.35} \approx 1 - 0.2592 \approx 0.7408 \] ### Final Answers Summary The dimensionless vessel dispersion number \( D^* \) is approximately \( 0.00222 \). The corrected overall chemical conversion for the first-order reaction is approximately \( 0.7408 \) or \( 74.08\% \).