Let's solve the problem step by step. ### **Given Data** - **Reaction:** \( A \rightarrow R \) (first-order, exothermic) - **Reactor:** PFR (Plug Flow Reactor), **adiabatic** - **Feed concentration:** \( C_{A0} = 2.5 \) mol/L - **Feed flowrate:** \( \dot{V}_0 = 360 \) L/h - **Initial temperature:** \( T_0 = 325 \) K - **\( \Delta H_{rxn} = -19000 \) J/mol (exothermic, so it's negative)** - **Specific heat capacities:** \( C_{pA} = 143.75 \) J/(mol·K) \( C_{pR} = 264.1 \) J/(mol·K) - **Rate constant:** \( k_A = 21 \exp(-1500/T) \) min\(^{-1}\) - **Required conversion:** \( X = 0.9 \) - **Density:** \( \rho = 0.85 \) g/cm\(^3\) (not needed unless volume or mass flowrate is required) - **Adiabatic process:** **No heat exchange (Q = 0)** --- ### **Step 1: Moles of A in Feed per Hour** \[ F_{A0} = C_{A0} \cdot \dot{V}_0 = 2.5~\text{mol/L} \times 360~\text{L/h} = 900~\text{mol/h} \] --- ### **Step 2: Adiabatic Energy Balance (Find Exit Temperature)** For **adiabatic PFR** (from your formula): \[ T = T_0 - \frac{\Delta H_{rxn}(T_0) \cdot X}{\sum_{i=1}^n \theta_i C_{p,i} + X \Delta C_p} \] #### **Find \(\sum \theta_i C_{p,i}\) and \(\Delta C_p\):** - Let’s assume the feed is pure A, so \(\theta_A = 1\), \(\theta_R = 0\) at inlet. - For 1 mole of A reacted, 1 mole of R formed (\(A \rightarrow R\)), so: - \(\theta_A = 1\) (initial moles of A per A in feed) - \(\theta_R = 0\) (no R in feed) - \(\Delta C_p = C_{pR} - C_{pA} = 264.1 - 143.75 = 120.35\) J/(mol·K) So, \[ \sum \theta_i C_{p,i} = 1 \times 143.75 + 0 \times 264.1 = 143.75~\text{J/(mol·K)} \] --- ### **Step 3: Calculate Exit Temperature** Plug values into the energy balance: \[ T = 325 - \frac{(-19000) \cdot 0.9}{143.75 + 0.9 \cdot 120.35} \] Calculate denominator: \[ 143.75 + 0.9 \times 120.35 = 143.75 + 108.315 = 252.065~\text{J/(mol·K)} \] Numerator: \[ (-19000) \times 0.9 = -17100 \] So, \[ T = 325 - \frac{-17100}{252.065} \] \[ T = 325 + 67.84 = 392.84~K \] --- ### **Step 4: Calculate Rate Constant at Exit Temperature** \[ k_A = 21 \exp\left(-\frac{1500}{T}\right) \] \[ T = 392.84~K \] \[ -1500/392.84 = -3.819 \] \[ \exp(-3.819) = 0.02199 \] \[ k_A = 21 \times 0.02199 = 0.4618~\text{min}^{-1} \] --- ### **Step 5: PFR Design Equation** For a **first-order, liquid-phase reaction, PFR:** \[ \frac{dX}{1-X} = \frac{k C_{A0}}{\dot{V}_0} dV \] \[ V = \int_0^X \frac{F_{A0}}{k C_{A0}} \frac{dX}{1-X} \] \[ V = \frac{F_{A0}}{k C_{A0}} \int_0^X \frac{dX}{1-X} = \frac{F_{A0}}{k C_{A0}} [-\ln(1-X)] \] Convert \(F_{A0}\) to **mol/min**: \[ F_{A0} = 900~\text{mol/h} = 15~\text{mol/min} \] \[ \dot{V}_0 = 360~\text{L/h} = 6~\text{L/min} \] So, \[ V = \frac{F_{A0}}{k C_{A0}} [-\ln(1-X)] \] Plug in numbers: \[ k = 0.4618~\text{min}^{-1} \] \[ C_{A0} = 2.5~\text{mol/L} \] \[ F_{A0} = 15~\text{mol/min} \] \[ X = 0.9 \] \[ -\ln(1-0.9) = -\ln(0.1) = 2.3026 \] \[ V = \frac{15}{0.4618 \times 2.5} \times 2.3026 \] \[ 0.4618 \times 2.5 = 1.1545 \] \[ \frac{15}{1.1545} = 12.995 \] \[ V = 12.995 \times 2.3026 = 29.92~\text{L} \] --- ## **Final Answers** ### **1. Exit Temperature** \[ T_{exit} = 392.8~K \] ### **2. Required Reactor Volume** \[ \boxed{V = 29.9~\text{L}} \] --- **If you need a step-by-step summary or further explanation, let me know!**