Let's break down the problem **step by step**: --- **Given Data:** - Reaction: **A ⇌ B** (elementary, reversible, liquid phase) - **Type:** CSTR (Continuous Stirred Tank Reactor), operated adiabatically - **Feed flow rate (\( v_0 \))** = **10 dm³/s** = **10 L/s** - **Feed concentration (\( C_{A0} \))** = **2 mol/L** - **Feed temperature (\( T_0 \))** = **300 K** - **Heat capacity (\( C_{pA} = C_{pB} \))** = **60 cal/(mol·K)** - **Reaction enthalpy (\( \Delta H^\circ_{rxn} \))** = **-10,000 cal/mol A** (exothermic) - **Required: 90% of equilibrium conversion** - **Find:** **Reactor volume (V)** --- ### **Step 1: Find the Equilibrium Conversion (\( X_{eq} \ ))** #### **1.1. Write the equilibrium for A ⇌ B** Let’s write the equilibrium expression: \[ K_c = \frac{C_{B,eq}}{C_{A,eq}} \] Let: - \( X_{eq} \) = equilibrium conversion of A Material balance at equilibrium: - \( C_{A,eq} = C_{A0}(1 - X_{eq}) \) - \( C_{B,eq} = C_{A0} X_{eq} \) So, \[ K_c = \frac{C_{A0} X_{eq}}{C_{A0}(1 - X_{eq})} = \frac{X_{eq}}{1 - X_{eq}} \] #### **1.2. Find \( K_c \) at equilibrium temperature** But **what is equilibrium temperature?** In an **adiabatic reactor**, the final temperature is determined by the extent of reaction. Let’s denote: - \( X \) = conversion - \( T \) = final temperature at conversion \( X \) --- ### **Step 2: Adiabatic Energy Balance** \[ \text{Heat generated by reaction} + \text{Heat required to increase temperature} = 0 \] Since it's adiabatic: \[ \sum n_i C_{pi} (T - T_0) = -\Delta H_{rxn} \cdot n_{A,reacted} \] But in a CSTR, for steady state, the **outlet** stream must be at temperature \( T \) corresponding to conversion \( X \). **Per mole of A fed:** - \( X \) moles of A react, producing X moles of B Total heat required to raise the mixture from \( T_0 \) to \( T \): \[ \text{Total heat capacity per mole of feed} = C_{pA}(1-X) + C_{pB} X \] But since \( C_{pA} = C_{pB} \), just use \( C_p \): \[ \text{Total heat required} = C_p (T - T_0) \] \[ \text{Heat generated by reaction} = -\Delta H_{rxn} \cdot X \] \[ C_p (T - T_0) = -\Delta H_{rxn} \cdot X \] \[ T = T_0 + \frac{-\Delta H_{rxn}}{C_p} X \] Plug in values: - \( \Delta H_{rxn} = -10,000 \) cal/mol (**exothermic**) - \( C_p = 60 \) cal/(mol·K) - \( T_0 = 300 \) K \[ T = 300 + \frac{-(-10,000)}{60} X \] \[ T = 300 + \frac{10,000}{60} X \] \[ T = 300 + 166.67 X \] --- ### **Step 3: Find \( K_c \) as a function of \( T \)** We need \( K_c \) at the final temperature. #### **Van't Hoff Equation:** \[ \ln \left( \frac{K_{c,T}}{K_{c,T_0}} \right) = -\frac{\Delta H^\circ_{rxn}}{R} \left( \frac{1}{T} - \frac{1}{T_0} \right) \] But we **don’t have \( K_{c,T_0} \)**. Let's proceed symbolically and see if it cancels later. --- ### **Step 4: 90% of Equilibrium Conversion** Let \( X_{target} = 0.9 X_{eq} \) --- ### **Step 5: Write the CSTR Design Equation** \[ \text{For a CSTR:} \qquad v_0 (C_{A0} - C_A) = r_A V \] \[ \Rightarrow V = \frac{v_0 (C_{A0} - C_A)}{-r_A} \] For elementary reversible reaction: \[ A \xrightleftharpoons[k_2]{k_1} B \] \[ r_A = -r = k_1 C_A - k_2 C_B \] But, since it's elementary and liquid phase: \[ r_A = -k_f C_A + k_r C_B \] But \( k_f \) and \( k_r \) can be related via \( K_c \): \[ K_c = \frac{k_f}{k_r} \implies k_r = \frac{k_f}{K_c} \] So: \[ -r_A = k_f C_A - k_r C_B = k_f (C_A - \frac{C_B}{K_c}) \] But \( C_A = C_{A0}(1-X) \), \( C_B = C_{A0} X \): \[ -r_A = k_f \left[ C_{A0}(1-X) - \frac{C_{A0} X}{K_c} \right] = k_f C_{A0} \left[ (1-X) - \frac{X}{K_c} \right] \] --- ### **Step 6: Plug In Everything** \[ V = \frac{v_0 (C_{A0} - C_A)}{-r_A} \] \[ C_{A0} - C_A = C_{A0} X \] So, \[ V = \frac{v_0 C_{A0} X}{k_f C_{A0} \left[ (1-X) - \frac{X}{K_c} \right]} = \frac{v_0 X}{k_f \left[ (1-X) - \frac{X}{K_c} \right]} \] --- ### **Step 7: Find \( X_{eq} \) Numerically** Recall: \[ K_c = \frac{X_{eq}}{1 - X_{eq}} \implies X_{eq} = \frac{K_c}{1 + K_c} \] So, at any temperature, \[ X_{eq}(T) = \frac{K_c(T)}{1 + K_c(T)} \] But at the target point, **target conversion** is \( X_{target} = 0.9 X_{eq} \). --- ### **Step 8: Relating \( X_{eq} \), \( K_c \), and \( T \)** Let’s proceed stepwise to solve numerically. Let’s define: Let’s denote \( X_{eq} = x \). \[ K_c = \frac{x}{1-x} \] But \( K_c \) is a function of \( T \): \[ \ln K_c = \ln K_{c,0} - \frac{\Delta H_{rxn}}{R} \left( \frac{1}{T} - \frac{1}{T_0} \right) \] But we **don’t have \( K_{c,0} \)**. Usually, in such problems, the value can be calculated from equilibrium data at given \( T_0 \), but since not provided, we can proceed with the symbolic answer, or make a reasonable assumption. But let's see what happens. Let’s set up all equations for the **final answer**: --- ## **Summary of Steps for Final Answer:** 1. **Energy Balance (Adiabatic):** \[ T = T_0 + \frac{-\Delta H_{rxn}}{C_p} X \] For the target conversion (\( X = X_{target} \)), this gives the outlet temperature. 2. **Equilibrium Constant:** \[ \ln K_c = \ln K_{c,0} - \frac{\Delta H_{rxn}}{R} \left( \frac{1}{T} - \frac{1}{T_0} \right) \] Where \( K_{c,0} \) is the equilibrium constant at \( T_0 \). If not given, you must assume or be provided its value. 3. **Equilibrium Conversion:** \[ X_{eq} = \frac{K_c}{1 + K_c} \] Target conversion: \( X_{target} = 0.9 X_{eq} \) 4. **CSTR Design Equation:** \[ V = \frac{v_0 X_{target}}{k_f \left[ (1 - X_{target}) - \frac{X_{target}}{K_c} \right]} \] Where \( k_f \) is the forward rate constant at \( T \): \[ k_f = k_{f,0} e^{-\frac{E}{R}\left(\frac{1}{T} - \frac{1}{T_0}\right)} \] (If \( k_{f,0} \) or \( E \) not given, leave as symbolic.) --- ## **If \( K_{c,0} \) and \( k_{f,0} \) are not given, you can write the answer as:** \[ \boxed{ T = 300 + 166.67 X_{target} \\ K_c = K_{c,0} \exp\left[ -\frac{-10,000}{1.987} \left( \frac{1}{T} - \frac{1}{300} \right) \right] \\ X_{eq} = \frac{K_c}{1 + K_c} \\ X_{target} = 0.9 X_{eq} \\ V = \frac{10 \times X_{target}}{k_f \left[ (1 - X_{target}) - \frac{X_{target}}{K_c} \right]} } \] Where: - \( C_{A0} = 2\, \text{mol/L} \) - \( v_0 = 10\, \text{L/s} \) - \( C_p = 60\, \text{cal/(mol·K)} \) - \( \Delta H_{rxn} = -10,000\, \text{cal/mol} \) - \( R = 1.987\, \text{cal/(mol·K)} \) - \( T_0 = 300\, \text{K} \) --- ### **If numerical values for \( K_{c,0} \) and \( k_{f,0} \) are provided, you can plug in and solve for \( V \).** --- ## **Final Step-by-Step Outline** ### **Step 1: Calculate \( X_{eq} \) at adiabatic temperature** Let \( X_{target} = 0.9 X_{eq} \). \[ T = 300 + 166.67 X_{target} \] \[ K_c = K_{c,0} \exp\left[\frac{10,000}{1.987}\left(\frac{1}{T} - \frac{1}{300}\right)\right] \] \[ X_{eq} = \frac{K_c}{1 + K_c} \] \[ X_{target} = 0.9 X_{eq} \] Repeat until converged (since \( X_{target} \) depends on \( K_c \) which depends on \( T \) which depends on \( X_{target} \)). ### **Step 2: Calculate reactor volume** \[ V = \frac{10 \times X_{target}}{k_f \left[ (1 - X_{target}) - \frac{X_{target}}{K_c} \right]} \] Where \( k_f \) is the forward rate constant at temperature \( T \). --- ## **Summary Table** | Step | Equation | |------|----------| | 1 | \( T = 300 + 166.67 X_{target} \) | | 2 | \( K_c = K_{c,0} \exp\left[\frac{10,000}{1.987}\left(\frac{1}{T} - \frac{1}{300}\right)\right] \) | | 3 | \( X_{eq} = \frac{K_c}{1 + K_c} \) | | 4 | \( X_{target} = 0.9 X_{eq} \) | | 5 | \( V = \frac{10 \times X_{target}}{k_f \left[ (1 - X_{target}) - \frac{X_{target}}{K_c} \right]} \) | --- ## **Final Answer (with explanation):** ### **The volume of CSTR required to achieve