Let's break down each part of the question step by step, providing explanations and final answers where possible. --- ## 1. **Stoichiometric Matrix and Gaussian Elimination** Given the matrix: \[ S - C - I = \begin{bmatrix} -1 & 1 \\ & -2 \\ 1 & \\ \end{bmatrix} \begin{bmatrix} n_A \\ n_B \\ \end{bmatrix} = \] **Step 1: Write the system of equations** \[ -1 n_A + 1 n_B = \\ n_A - 2 n_B = \\ 1 n_A + n_B = \\ \] **Step 2: Write in matrix form and reduce** The matrix is: \[ \begin{bmatrix} -1 & 1 \\ & -2 \\ 1 & \\ \end{bmatrix} \] **Row 2** gives: \( -2 n_B = \implies n_B = \) **Row 1**: \( -n_A + n_B = \implies n_A = n_B \implies n_A = \) **Row 3**: \( n_A = \) **Step 3: Rank** There are 3 equations, 2 unknowns. The matrix's rank is 2 (since the first two rows are linearly independent). Therefore, the minimum number of independent reactions needed is 2. **Final Answer:** **Rank = 2** --- ## 2. **Graphing Selectivity \( S_{P/Q} \) vs. Conversion of \( A \)** **Explanation:** For a parallel reaction: \[ A \xrightarrow{k_1} P \\ A \xrightarrow{k_2} Q \\ \] Selectivity \( S_{P/Q} = \frac{r_P}{r_Q} = \frac{k_1}{k_2} \frac{[A]}{[A]} = \frac{k_1}{k_2} \) But as conversion increases, \([A]\) decreases, and selectivity changes if the reactions are not first-order. **Graph Features:** - X-axis: Conversion of \(A\) ( to 1) - Y-axis: \(S_{P/Q}\) - Higher \(k_1/k_2\) ratio gives higher selectivity to P. - Maintaining high \(C_A\) (low conversion) generally favors selectivity to P if P’s reaction is faster. **Plot:** Draw a decreasing curve if Q is higher order, or flat if both are first order. --- ## 3. **Maximum Yield of P** For parallel reactions: \[ Y_P = \frac{k_1}{k_1 + k_2} (1 - e^{-(k_1 + k_2)t}) \] Maximum yield is obtained as \( t \to \infty \): \[ Y_{P,\text{max}} = \frac{k_1}{k_1 + k_2} \] **Final Answer:** \[ Y_{P,\text{max}} = \frac{k_1}{k_1 + k_2} \] --- ## 4. **Graph Internal Effectiveness \( \eta \) vs Thiele Modulus \( \phi \)** **Expression:** \[ \eta = \frac{\int_V r_A(C, T) dV}{r_A(C_s, T_s) V_p} \] The effectiveness factor decreases as \(\phi\) increases (diffusion limits reaction). For non-isothermal, multiple steady states occur at high \(\gamma\) and \(\beta\). **Plot Features:** - X-axis: \(\phi\) - Y-axis: \(\eta\) - Show decreasing trend, possible S-shape (multiple steady states). --- ## 5. **Mean Residence Time & Variance from Pulse Data** Given: \[ \begin{array}{ccc} t\,(\text{min}) & 2 & 4 & 6 & 8 & 10 \\ C(t)\,(\text{g}/L) & .1 & 1.2 & 2.8 & 1.3 & .2 \\ \end{array} \] ### **Step 1: Normalize \(C(t)\) to get \(E(t)\):** Total area under curve = \(.1 + 1.2 + 2.8 + 1.3 + .2 = 5.6\) So, \[ E(t) = \frac{C(t)}{5.6} \] ### **Step 2: Mean Residence Time \(\bar{t}\)** \[ \bar{t} = \sum t_i E(t_i) \] \[ = 2 \times \frac{.1}{5.6} + 4 \times \frac{1.2}{5.6} + 6 \times \frac{2.8}{5.6} + 8 \times \frac{1.3}{5.6} + 10 \times \frac{.2}{5.6} \] Calculate: \[ = 2 \times .0179 + 4 \times .2143 + 6 \times .5 + 8 \times .2321 + 10 \times .0357 \] \[ = .0357 + .8571 + 3 + 1.8571 + .3571 = 6.107 \] ### **Step 3: Variance \(\sigma^2\)** \[ \sigma^2 = \sum (t_i - \bar{t})^2 E(t_i) \] Calculate each term: - For \(t_1 = 2\): \((2-6.107)^2 \times .0179 = 16.882 \times .0179 = .302\) - For \(t_2 = 4\): \((4-6.107)^2 \times .2143 = 4.442 \times .2143 = .952\) - For \(t_3 = 6\): \((6-6.107)^2 \times .5 = .0115 \times .5 = .006\) - For \(t_4 = 8\): \((8-6.107)^2 \times .2321 = 3.59 \times .2321 = .833\) - For \(t_5 = 10\): \((10-6.107)^2 \times .0357 = 15.123 \times .0357 = .540\) Sum: \(.302 + .952 + .006 + .833 + .540 = 2.633\) **Final Answers:** \[ \bar{t} \approx 6.11 \text{ min}, \quad \sigma^2 \approx 2.63 \text{ min}^2 \] --- ## 6. **Dispersion Number \(\frac{D}{uL}\)** \[ \frac{D}{uL} = \frac{\sigma^2}{2 \bar{t}^2} \] \[ = \frac{2.63}{2 \times (6.11)^2} = \frac{2.63}{74.68} \approx .035 \] **Final Answer:** \[ \frac{D}{uL} \approx .035 \] --- ## 7. **Catalyst Weight for 40% Conversion in Adiabatic Packed Bed** Given: \[ \frac{dX}{dW} = \frac{k(T_{in} + \Delta T_{ad} X)}{F_{A}} (1-X) \left(\frac{T_{in}}{T_{in} + \Delta T_{ad} X}\right)^2 \] **Process:** - Integrate for \(X\) from to .4. - Requires numerical integration given \(k(T)\) (usually Arrhenius). **Final:** Set up for numerical solution; cannot solve analytically without more data. --- ## 8. **Hysteresis Loop for Autothermal Reactor** **Sketch:** - X-axis: \(T\) - Y-axis: Heat generated (\(G(T)\)) and removed (\(R(T)\)) - Sigmoid curve for \(G(T)\), linear for \(R(T)\) - Tangency points = ignition/extinction points. --- ## 9. **Minimum Ignition Temperature** Set: \[ G(T) = R(T) \] and at the ignition point, \[ \frac{dG}{dT} = \frac{dR}{dT} \] Solve these two equations simultaneously for \(T_{ign}\). --- ### **Summary Table** | Part | Final Answer/Approach | |------|----------------------| | 1 | Rank = 2 | | 2 | Plot selectivity vs. conversion (see explanation) | | 3 | \( Y_{P,\text{max}} = \frac{k_1}{k_1 + k_2} \) | | 4 | Plot \(\eta\) vs. \(\phi\) (see explanation) | | 5 | \(\bar{t} \approx 6.11\,\text{min}, \sigma^2 \approx 2.63\,\text{min}^2\) | | 6 | \( \frac{D}{uL} \approx .035 \) | | 7 | Numerical integration of ODE for \(W\) | | 8 | Hysteresis sketch: S-curve and straight line, tangency points are ignition/extinction | | 9 | Solve \( G(T)=R(T) \) and \( dG/dT=dR/dT \) simultaneously for \( T_{ign} \) | If you need more specific calculations or plots, let me know!