Let's break down and solve the problem step by step as described in your image. --- # **Step 1: Problem Summary** A **first order irreversible reaction**: \[ A \rightarrow B \] **In a plug flow reactor (PFR)**, given: - \( F_{A} = 5 \text{ kmol/hr} \) - \( C_{A} = 1.5 \text{ kmol/m}^3 \) - \( k = .8 \text{ hr}^{-1} \) The rate equation: \[ -r_A = kC_A \] \[ C_A = C_{A}(1-x) \] where \( x \) is the conversion. --- # **Step 2: Reactor Volume Required** For a PFR, the volume required is: \[ V = \int_^x \frac{F_{A}}{-r_A} \, dx \] Substitute the given values: \[ -r_A = kC_A = k C_{A}(1-x) \] So, \[ V = \int_^x \frac{F_{A}}{k C_{A}(1-x)} dx \] Take constants outside the integral: \[ V = \frac{F_{A}}{k C_{A}} \int_^x \frac{1}{1-x} dx \] The integral is: \[ \int_^x \frac{1}{1-x} dx = -\ln(1-x) \bigg|_^x = -\ln(1-x) + \ln(1-) = -\ln(1-x) \] So, \[ V = \frac{F_{A}}{k C_{A}} [-\ln(1-x)] \] --- # **Step 3: Plug in the Numbers** \[ F_{A} = 5 \text{ kmol/hr} \] \[ k = .8 \text{ hr}^{-1} \] \[ C_{A} = 1.5 \text{ kmol/m}^3 \] So, \[ V = \frac{5}{.8 \times 1.5} [ -\ln(1-x) ] \] \[ V = \frac{5}{1.2} [ -\ln(1-x) ] \] \[ V = 4.167 [ -\ln(1-x) ] \text{ m}^3 \] --- # **Step 4: Space Time** Space time, \( \tau \): \[ \tau = \frac{V}{V_} \] where \[ V_ = \frac{F_{A}}{C_{A}} \] \[ V_ = \frac{5}{1.5} = 3.333 \text{ m}^3 \] So, \[ \tau = \frac{V}{3.333} \] --- # **Final Answer** ## **1. Reactor Volume Required:** \[ \boxed{ V = 4.167 \left[ -\ln(1-x) \right] \text{ m}^3 } \] where \( x \) is the desired conversion. ## **2. Space Time:** \[ \boxed{ \tau = \frac{V}{3.333} } \] where \( V \) is the reactor volume found above. --- ### **To find the volume for a specific conversion** (\( x \)), just substitute the value of \( x \) in the formula. If you need \( V \) for a particular \( x \), let me know!