Let's solve each part step by step with explanations: **1. Calculate the equilibrium constant \( K_c \) if \(|B|_{eq} = 8\) and \(|A|_{eq} = 2\):** Given the relation: \[ K_c = \frac{\text{Product}_{eq}}{\text{Reactant}_{eq}} \] Assume the reaction is \(A \rightleftharpoons B\): \[ K_c = \frac{[B]_{eq}}{[A]_{eq}} \] Plug in the values: \[ K_c = \frac{8}{2} = 4 \] So, the equilibrium constant \( K_c \) is **4**. --- **2. Calculate the residence time \(\tau\) for \(V = 100\) L and \(v_ = 5\) L/min:** The formula for residence time is: \[ \tau = \frac{V}{v_} \] Plug in the given values: \[ \tau = \frac{100}{5} = 20 \text{ min} \] So, the residence time \(\tau\) is **20 minutes**. --- **3. Linearization of the Arrhenius Equation and Identifying Intercept:** The linearized Arrhenius equation is: \[ \ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right) \] When you plot \(\ln k\) versus \(1/T\), the intercept on the y-axis is \(\ln A\). --- **4. Determine \(\ln A\) if the intercept is 5.:** Given that the intercept is 5., then: \[ \ln A = 5. \] --- **Final Answers:** 1. The equilibrium constant \(K_c\) is **4**. 2. The residence time \(\tau\) is **20 minutes**. 3. The y-intercept in the Arrhenius plot is \(\ln A\). 4. The value of \(\ln A\) is **5.**.