Let's solve the problem step-by-step: --- ### **Given Data** - **Reaction:** \(A \rightleftharpoons B\) (elementary, reversible, liquid phase) - **Type:** CSTR (Continuous Stirred Tank Reactor), adiabatic operation - **Feed flow rate (\(F_{in}\)):** \(10\, \text{dm}^3/\text{s}\) - **Feed concentration of A (\(C_{A0}\)):** \(2\, \text{mol/dm}^3\) - **Feed temperature (\(T_0\)):** \(300\, \text{K}\) - **Heat capacity (\(C_p\)):** \(60\, \text{cal/(mol K)}\) (for both A and B, assume liquid mixtures, so \(\bar{C}_p\) is fine) - **Standard enthalpy of reaction (\(\Delta H^\circ\)):** \(-10{,}000\, \text{cal/mol A}\) - **Required:** Volume of CSTR to achieve 90% of equilibrium conversion, adiabatic --- ### **Step 1: Find Equilibrium Conversion (\(X_{eq}\)) at Feed Conditions** #### **1.1. Write the Equilibrium Constant Expression** For the reaction \(A \rightleftharpoons B\): \[ K_c = \frac{C_{B,eq}}{C_{A,eq}} \] Let initial moles of A (per dm³) = \(C_{A0}\), and suppose at equilibrium \(X_{eq}\) is the conversion. \[ \begin{align*} C_{A,eq} &= C_{A0}(1 - X_{eq}) \\ C_{B,eq} &= C_{A0} X_{eq} \\ K_c &= \frac{C_{B,eq}}{C_{A,eq}} = \frac{X_{eq}}{1 - X_{eq}} \end{align*} \] #### **1.2. Find \(K_c\) at \(T_0 = 300\, \text{K}\)** But we don't have \(K_c\) directly. However, we can relate it to the standard Gibbs energy change: \[ \Delta G^\circ = -RT \ln K_c \] But without \(\Delta G^\circ\), we can use the van't Hoff equation, given \(\Delta H^\circ\). But since we don't have \(K_{c,ref}\) at any reference temperature, **we cannot find \(X_{eq}\) numerically** unless more data is given. However, let's **express the answer in terms of \(K_c\)**, and carry it through. --- ### **Step 2: Find the Target Conversion** You are to achieve **90% of equilibrium conversion**: \[ X = 0.9 X_{eq} \] --- ### **Step 3: Adiabatic Energy Balance** For adiabatic reactors, the temperature rise is due to reaction heat: \[ \text{Moles of A reacted per (dm³)} = C_{A0}(X) \] \[ \text{Energy released per (dm³)} = C_{A0} X \cdot (-\Delta H^\circ) \] This energy raises the temperature: \[ C_{A0} X \cdot (-\Delta H^\circ) = (C_{A0} (1 - X) + C_{A0} X) C_p (T - T_0) \] But since total moles remain same (1 mol A consumed, 1 mol B formed), total moles = \(C_{A0}\) per dm³. So, \[ C_{A0} X (-\Delta H^\circ) = C_{A0} C_p (T - T_0) \] \[ X (-\Delta H^\circ) = C_p (T - T_0) \] \[ T = T_0 + \frac{-\Delta H^\circ}{C_p} X \] Plug in numbers: - \(-\Delta H^\circ = 10,\!000\, \text{cal/mol}\) - \(C_p = 60\, \text{cal/(mol K)}\) - \(T_0 = 300\, \text{K}\) \[ T = 300 + \frac{10,\!000}{60} X \] \[ T = 300 + 166.67 X \] --- ### **Step 4: Find \(K_c\) at Operating Temperature** \[ K_c = \frac{X_{eq}}{1 - X_{eq}} \] But the equilibrium constant depends on temperature. If we had \(K_{c,0}\) at \(T_0\), we could use van't Hoff: \[ \ln \frac{K_{c,T}}{K_{c,T_0}} = -\frac{\Delta H^\circ}{R} \left( \frac{1}{T} - \frac{1}{T_0} \right) \] Since we don't have a value for \(K_{c,T_0}\), let's **proceed symbolically**. --- ### **Step 5: Rate Law** For an elementary, reversible reaction: \[ r_A = -k_1 C_A + k_{-1} C_B \] But: \[ k_{-1} = \frac{k_1}{K_c} \] So: \[ r_A = -k_1 C_A + \frac{k_1}{K_c} C_B = -k_1 \left( C_A - \frac{C_B}{K_c} \right) \] But since \(C_A = C_{A0}(1 - X)\), \(C_B = C_{A0} X\): \[ r_A = -k_1 C_{A0} \left[ (1 - X) - \frac{X}{K_c} \right] \] \[ r_A = -k_1 C_{A0} \left[ 1 - X - \frac{X}{K_c} \right] \] --- ### **Step 6: CSTR Mole Balance** For a CSTR: \[ F_{in} C_{A0} - F_{out} C_A + r_A V = 0 \] For constant flow rate (\(F_{in} = F_{out}\)), at steady-state: \[ F (C_{A0} - C_A) = -r_A V \] But \(C_{A0} - C_A = C_{A0} X\): \[ F C_{A0} X = -r_A V \] \[ V = \frac{F C_{A0} X}{-r_A} \] Plug in the rate: \[ V = \frac{F C_{A0} X}{k_1 C_{A0} \left[ 1 - X - \frac{X}{K_c} \right]} \] \[ V = \frac{F X}{k_1 \left[ 1 - X - \frac{X}{K_c} \right]} \] --- ### **Step 7: Find \(k_1\) at Operating Temperature** The forward rate constant: \[ k_1 = k_{1,0} \exp \left( -\frac{E_a}{RT} \right) \] But **we are not given \(E_a\) or \(k_{1,0}\)**! **Therefore, unless more data is given, you must leave the answer in terms of \(k_1\) at the operating temperature.** --- ### **Step 8: Final Formula** - \(X = 0.9 X_{eq}\) - \(T = 300 + 166.67 X\) - \(K_c(T) = \frac{X_{eq}}{1 - X_{eq}}\) - \(V = \frac{F X}{k_1(T) [1 - X - X/K_c(T)]}\) #### **If \(K_c\) is large (forward reaction favored, \(X_{eq} \approx 1\)):** \[ X_{eq} \approx 1, \; X = 0.9 \] \[ K_c \gg 1, \quad \frac{X}{K_c} \to 0 \] \[ V = \frac{F X}{k_1(T)(1 - X)} \] --- ### **Step 9: Plug Numbers for a Hypothetical Case** Assume \(X_{eq} = 0.95\) (just for illustration): \[ X = 0.9 \times 0.95 = 0.855 \] \[ T = 300 + 166.67 \times 0.855 = 442.5\, \text{K} \] Suppose at this \(T\), \(K_c = \frac{0.95}{0.05} = 19\) So, \[ 1 - X - \frac{X}{K_c} = 1 - 0.855 - \frac{0.855}{19} = 0.145 - 0.045 = 0.10 \] So, \[ V = \frac{10 \times 0.855}{k_1(442.5) \times 0.10} = \frac{8.55}{0.10 k_1(442.5)} = \frac{85.5}{k_1(442.5)} \] --- ## **Final Answer** ### **General Solution (in terms of knowns):** \[ \boxed{ V = \frac{F X}{k_1(T)\left[1 - X - \frac{X}{K_c(T)}\right]} } \] where: - \(F = 10\, \text{dm}^3/\text{s}\) - \(X = 0.9 X_{eq}\) - \(T = 300 + \dfrac{10,\!000}{60} X\) - \(K_c(T) = \dfrac{X_{eq}}{1 - X_{eq}}\) at temperature \(T\) - \(k_1(T)\) is the forward rate constant at temperature \(T\) **To compute a numerical answer, you need either \(X_{eq}\) or \(K_c\) at the relevant temperature, and \(k_1\) at that temperature. If you have those, plug them in as shown above.** --- ### **If You Assume High Equilibrium Conversion (\(X_{eq} \approx 1\)):** \[ V \approx \frac{F X}{k_1(T) (1 - X)} \] with \(X = 0.9\). --- #### **Summary Table** | Step | Equation/Expression | |---------------------|-------------------------------------------------------| | Adiabatic \(T\) | \(T = T_0 + \frac{-\Delta H^\circ}{C_p} X\) | | \(K_c\) | \(K_c = \frac{X_{eq}}{1-X_{eq}}\) at \(T\) | | CSTR balance | \(V = \frac{F X}{k_1(T)\left[1 - X - \frac{X}{K_c}\right]}\) | | Target conversion | \(X = 0.9 X_{eq}\) | --- **If more data is given (e.g., \(k_1\) or \(K_c\)), you can compute a numerical value. Otherwise, the above is the stepwise, general solution.**