Let's solve each part step by step: --- ## **(a) Adiabatic Operation: Equilibrium Temperature Calculation** ### **Given Data** - Reaction: \(A \rightleftharpoons B\) - Feed: \(298\,K\), equimolar \(A\) and inert - \(\Delta H_{r,298K} = -80{,}000\,\text{J/mol}\) (exothermic) - \(C_{p,A} = 50\,\text{J/mol K}\) - \(C_{p,B} = 50\,\text{J/mol K}\) - \(C_{p,\text{inert}} = 200\,\text{J/mol K}\) - \(K_{298K} = 26750\) **Objective:** Show a graph for adiabatic operation reaching equilibrium at **440 K**. --- ### **Step 1: Adiabatic Energy Balance** For an adiabatic reactor, the energy released/absorbed by reaction increases/decreases the temperature. For \(x\) mol conversion of \(A\): \[ \text{Total } C_p = C_{p,A} + C_{p,\text{inert}} = 50 + 200 = 250\,\text{J/mol K} \] (Initial: 1 mol A, 1 mol inert) Number of moles after conversion \(x\) (for 1 mol A fed): - Moles A left: \(1 - x\) - Moles B formed: \(x\) - Inert: 1 mol Total \(C_p\) after reaction: \[ C_{p,\text{total}} = (1-x)C_{p,A} + x C_{p,B} + 1 C_{p,\text{inert}} \] But since \(C_{p,A} = C_{p,B}\), \(C_{p,\text{total}} = 50 + 200 = 250\,\text{J/mol K}\) (constant). --- #### **Adiabatic Temperature Rise** \[ \Delta T = \frac{-\Delta H_{r,298K} \cdot x}{\Sigma n_i C_{p,i}} \] Let \(x_{eq}\) be the equilibrium conversion at \(440\,K\): \[ T_{final} = 298 + \frac{-\Delta H_{r,298K} \cdot x_{eq}}{C_{p,\text{total}}} \] --- ### **Step 2: Equilibrium Constant as a Function of T** \[ \Delta G^\circ = -RT \ln K \] \[ \ln K = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T}\right) + \frac{\Delta S^\circ}{R} \] But, since we know \(K_{298K}\), we can use the van't Hoff equation: \[ \ln\left(\frac{K_T}{K_{298}}\right) = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T} - \frac{1}{298}\right) \] Given: - \(K_{298} = 26750\) - \(\Delta H^\circ = -80,000\,\text{J/mol}\) - \(R = 8.314\,\text{J/mol K}\) --- ### **Step 3: Expression for Equilibrium Conversion** For an equimolar feed of A and inert, and a simple \(A \rightarrow B\): \[ A_{initial} = 1,\; I_{initial} = 1 \] At conversion \(X\): \[ A = 1 - X,\; B = X,\; I = 1 \] Mole fractions: \[ y_A = \frac{1-X}{2},\quad y_B = \frac{X}{2},\quad y_I = \frac{1}{2} \] Equilibrium constant (for liquid, \(K = \frac{a_B}{a_A}\), assume ideal): \[ K = \frac{[B]}{[A]} = \frac{X}{1-X} \] --- ### **Step 4: At Equilibrium, Find \(X_{eq}\) at 440 K** First, compute \(K_{440}\): \[ \ln\left(\frac{K_{440}}{26750}\right) = -\frac{-80,000}{8.314} \left(\frac{1}{440} - \frac{1}{298}\right) \] \[ = 9625.2 \left( \frac{1}{440} - \frac{1}{298} \right) = 9625.2 \left(.002273 - .003356 \right) = 9625.2 \times (-.001083) = -10.43 \] \[ \ln\left(\frac{K_{440}}{26750}\right) = -10.43 \] \[ \frac{K_{440}}{26750} = e^{-10.43} = 2.94 \times 10^{-5} \] \[ K_{440} = 26750 \times 2.94 \times 10^{-5} \approx .786 \] At equilibrium: \[ K_{440} = \frac{X_{eq}}{1-X_{eq}} \] \[ .786 = \frac{X_{eq}}{1-X_{eq}} \implies X_{eq} = .786(1-X_{eq}) \implies X_{eq} = .786 - .786 X_{eq} \implies X_{eq}(1 + .786) = .786 \implies X_{eq} = \frac{.786}{1.786} = .44 \] --- ### **Step 5: Adiabatic Temperature Rise** Verify if this matches the adiabatic calculation: \[ T = 298 + \frac{-\Delta H_r x_{eq}}{C_{p,total}} = 298 + \frac{80,000 \times .44}{250} = 298 + 140.8 = 438.8\,K \] This is very close to 440 K. --- #### **Adiabatic Path Graph** - **X-axis:** Conversion (\(X\)), from to \(X_{eq}\) - **Y-axis:** Temperature (\(T\)), from 298 K to 440 K The graph should be a straight line (since \(\Delta H\) and \(C_p\) are constant): \[ T = 298 + \frac{80,000}{250} X = 298 + 320 X \] At \(X=\): \(T=298\) K At \(X=.44\): \(T=438.8 \approx 440\) K --- ## **(b) Interstage Cooling to Double Conversion** **Target:** Achieve \(2 \times .44 = .88\) conversion using interstage cooling. ### **Step 1: Equilibrium Conversion at Lower Temperature** At 298 K: \[ K_{298} = \frac{X_{eq}}{1-X_{eq}} = 26750 \] \[ X_{eq,298} = \frac{26750}{1+26750} \approx .99996 \] At 348 K: \[ \ln\left(\frac{K_{348}}{26750}\right) = -\frac{-80,000}{8.314} \left(\frac{1}{348} - \frac{1}{298}\right) = 9625.2 \times (.002874 - .003356) = 9625.2 \times (-.000482) = -4.64 \] \[ K_{348} = 26750 \times e^{-4.64} = 26750 \times .0097 = 259.48 \] \[ X_{eq,348} = \frac{259.48}{1+259.48} = .996 \] So, at lower temperatures, conversion is nearly complete. --- ### **Step 2: Number of Reactors Needed** Each reactor (adiabatic) raises the temperature, so after each stage, cool back between 298 K and 348 K. Let \(X_1\) = conversion in first reactor (from 298 K to 348 K): \[ T = 298 + \frac{80,000 X_1}{250} = 348 \implies 348 - 298 = 320 X_1 \implies 50 = 320 X_1 \implies X_1 = \frac{50}{320} = .156 \] After first reactor, conversion is .156. Cool back to 298 K, treat unreacted A as new feed (A left = 1-.156 = .844). Repeat: Second reactor: feed = .844 mol A, .156 mol B, 1 mol inert; conversion in reactor is again .156 of remaining A. **General pattern:** \[ \text{Remaining A after n stages: } (1 - X_1)^n \] \[ \text{Total conversion after n stages: } 1 - (1 - X_1)^n \] Set this equal to .88 to solve for \(n\): \[ 1 - (1 - .156)^n = .88 \implies (1 - .156)^n = .12 \implies (.844)^n = .12 \implies n \ln .844 = \ln .12 \implies n = \frac{\ln .12}{\ln .844} \] \[ \ln .12 = -2.1203,\quad \ln .844 = -.1693 \] \[ n = \frac{-2.1203}{-.1693} = 12.53 \] So, **13 reactors** (rounded up) are needed to achieve at least 88% conversion. --- ## **(c) Alternative Method to Double Equilibrium Conversion** **Possible options:** - **Remove product B continuously** (shifts equilibrium to the right). - **Increase pressure** (not effective for liquid-phase, but for gas-phase reactions with fewer moles, it helps). - **Use a different solvent** (to change activity coefficients if non-ideal). - **Change catalyst** to favor forward reaction if possible. - **Apply a membrane reactor** to selectively remove B. **Best option here:** **Remove product B as it forms, e.g., using a membrane reactor or in situ extraction.** --- # **Summary Table** | Part | Step | Result/Formula | Value/Explanation | |------|------|---------------|------------------| | (a) | Adiabatic Eq. | \(T = 298 + 320X\) | X from to .44, T from 298 K to 440 K | | (a) | Equilibrium at 440 K | \(K_{440} = .786\), \(X_{eq} = .44\) | | | (b) | 1 stage conversion | \(X_1 = .156\) (from 298 K to 348 K) | | | (b) | No. of reactors | \(n = \frac{\ln .12}{\ln .844} = 12.53\) | **13 reactors** required | | (c) | Alternative | Remove B as formed | e.g., membrane reactor | --- # **Final Answers** ### **(a)** - Adiabatic plot: \(T = 298 + 320X\) from \(X=\) to \(X=.44\), \(T=298\) K to \(T=440\) K. - Equilibrium conversion at 440 K is **.44**. ### **(b)** - To double conversion (to .88) using interstage cooling (298–348 K), **13 reactors** are needed. ### **(c)** - **Alternative method:** Remove product B as it forms (e.g., via membrane reactor or in situ extraction) to shift equilibrium and increase conversion.