Let's break down your problem **step by step**: --- ## **Given Data** - **Reaction:** \( A \rightarrow R \) (series: CSTR → PFR) - **Rate law:** \( -r_A = .02C_A - .01C_R \) - **Volumes:** \( V_{CSTR} = V_{PFR} = 200\,L \) - **Initial concentration:** \( C_{A} = .1\,mol/L \) - **Flow rate:** \( Q = 100\,L/min \) We need to **calculate the final conversion** and **discuss if this configuration is possible**. --- ## **Step 1: CSTR Material Balance** For a **CSTR**: \[ F_{A} - F_{A1} + r_{A1}V = \] Where: - \( F_{A} = Q C_{A} \) (inlet molar flow of A) - \( F_{A1} = Q C_{A1} \) (outlet molar flow of A) - \( r_{A1} = .02C_{A1} - .01C_{R1} \) (rate at outlet) Since \( C_{R} = \) (no R in inlet), and in CSTR, outlet conc. is uniform: - At the outlet: \( C_{A1}, C_{R1} \) Also, for a single reaction: \[ C_{A} - C_{A1} = C_{R1} \implies C_{R1} = C_{A} - C_{A1} \] Plug into the CSTR balance: \[ Q(C_{A} - C_{A1}) + (.02C_{A1} - .01C_{R1})V = \] Plug \( Q = 100\,L/min \), \( V = 200\,L \): \[ 100(C_{A} - C_{A1}) + 200(.02C_{A1} - .01C_{R1}) = \] \[ 100(C_{A} - C_{A1}) + 40C_{A1} - 20C_{R1} = \] \[ 100C_{A} - 100C_{A1} + 40C_{A1} - 20C_{R1} = \] \[ 100C_{A} - 60C_{A1} - 20C_{R1} = \] Substitute \( C_{R1} = C_{A} - C_{A1} \): \[ 100C_{A} - 60C_{A1} - 20(C_{A} - C_{A1}) = \] \[ 100C_{A} - 60C_{A1} - 20C_{A} + 20C_{A1} = \] \[ (100C_{A} - 20C_{A}) + (-60C_{A1} + 20C_{A1}) = \] \[ 80C_{A} - 40C_{A1} = \] \[ C_{A1} = 2C_{A} \] But that's not physically possible since \( C_{A1} < C_{A} \) (it can't increase). **Let's check calculation:** Let's write it more clearly: \[ 100C_{A} - 60C_{A1} - 20C_{R1} = \] \[ 100C_{A} - 60C_{A1} - 20(C_{A} - C_{A1}) = \] \[ 100C_{A} - 60C_{A1} - 20C_{A} + 20C_{A1} = \] \[ 100C_{A} - 20C_{A} = 80C_{A} \] \[ -60C_{A1} + 20C_{A1} = -40C_{A1} \] So: \[ 80C_{A} - 40C_{A1} = \] \[ 2C_{A} = C_{A1} \] But since \( C_{A} = .1 \), \( C_{A1} = .2 \) which is not possible. This suggests **negative conversion** (impossible). --- ## **Physical Meaning and Feasibility** If the rate law is \( -r_A = .02C_A - .01C_R \), and \( C_{R} \) builds up, the reaction can **reverse** if \( .01C_R > .02C_A \) (i.e., the back reaction dominates). Given \( C_{A} = .1 \), let's test what happens at the PFR. --- ## **Step 2: PFR Calculation** The PFR follows: \[ \frac{dF_A}{dV} = r_A \] \[ \frac{dC_A}{dV} = \frac{r_A}{Q} \] But if the CSTR already gives us an impossible or unphysical result, the setup is **not feasible**. Let's try a direct mass balance for the CSTR: --- ### **CSTR Only, Check Maximum Conversion** Let \( X \) be conversion after CSTR, \[ C_{A1} = C_{A}(1 - X_1) \] \[ C_{R1} = C_{A}X_1 \] Plug into the CSTR balance: \[ Q(C_{A} - C_{A1}) = r_{A1} V \] \[ Q(C_{A} - C_{A1}) = (.02C_{A1} - .01C_{R1})V \] \[ Q(C_{A} - C_{A1}) = (.02C_{A1} - .01C_{R1})V \] Plug in values: \[ 100(C_{A} - C_{A1}) = (.02C_{A1} - .01C_{R1})200 \] Substitute \( C_{A} = .1 \), \( C_{A1} = .1 - X_1 \cdot .1 \), \( C_{R1} = X_1 \cdot .1 \): Let’s let \( C_{A1} = y \), so \( C_{R1} = .1 - y \): \[ 100(.1 - y) = [.02y - .01(.1 - y)]200 \] \[ 100(.1 - y) = [.02y - .001 + .01y]200 \] \[ 100(.1 - y) = (.03y - .001)200 \] \[ 100(.1 - y) = 60y - 2 \] \[ 10 - 100y = 60y - 2 \] \[ 10 + 2 = 60y + 100y \] \[ 12 = 160y \] \[ y = \frac{12}{160} = .075 \] So, \[ C_{A1} = .075\,mol/L \] \[ C_{R1} = .1 - .075 = .025\,mol/L \] --- ## **Step 3: PFR Calculation** For the PFR: \[ \frac{dC_A}{dV} = \frac{r_A}{Q} \] \[ \frac{dC_A}{dV} = \frac{.02C_A - .01C_R}{100} \] But, as the reaction proceeds, \( C_A \) decreases, \( C_R \) increases. Let’s use the **stoichiometry**: For every decrease in \( C_A \), \( C_R \) increases by same amount: Let \( C_{A2} \) and \( C_{R2} \) be at the PFR outlet. \[ C_{A2} = C_{A1} - \Delta \] \[ C_{R2} = C_{R1} + \Delta \] The differential for \( C_A \): \[ \frac{dC_A}{dV} = \frac{.02C_A - .01C_R}{100} \] But as \( C_A \) decreases and \( C_R \) increases, the rate will eventually reach zero (equilibrium): Set \( .02C_A - .01C_R = \implies 2C_A = C_R \) Use initial values from CSTR outlet: From above, \( C_{A1} = .075 \), \( C_{R1} = .025 \) Let’s try to integrate: Let \( x \) be the amount of A converted in the PFR, so: \[ C_A = .075 - x \] \[ C_R = .025 + x \] The rate: \[ \frac{dx}{dV} = -\frac{.02C_A - .01C_R}{100} \] \[ \frac{dx}{dV} = -\frac{.02(.075 - x) - .01(.025 + x)}{100} \] \[ = -\frac{.02 \times .075 - .02x - .01 \times .025 - .01x}{100} \] \[ = -\frac{.0015 - .02x - .00025 - .01x}{100} \] \[ = -\frac{.00125 - .03x}{100} \] But since \( dx \) is positive as A is consumed: \[ \frac{dx}{dV} = \frac{.00125 - .03x}{100} \] \[ 100 dx = (.00125 - .03x)dV \] \[ \frac{dx}{.00125 - .03x} = \frac{dV}{100} \] Integrate from \( x = \) to \( x = x_f \), \( V = \) to \( V = 200 \): \[ \int_{}^{x_f} \frac{dx}{.00125 - .03x} = \int_{}^{200} \frac{dV}{100} \] \[ \left[ -\frac{1}{.03} \ln(.00125 - .03x) \right]_^{x_f} = \left[ \frac{V}{100} \right]_^{200} \] \[ -\frac{1}{.03} \left[ \ln(.00125 - .03x_f) - \ln(.00125) \right] = 20 \] \[ -\frac{1}{.03} \ln\left( \frac{.00125 - .03x_f}{.00125} \right) = 20 \] \[ \ln\left( \frac{.00125 - .03x_f}{.00125} \right) = -.03 \times 20 = -.6 \] \[ \frac{.00125 - .03x_f}{.00125} = e^{-.6} = .5488 \] \[ .00125 - .03x_f = .00125 \times .5488 = .000686 \] \[ .03x_f = .00125 - .000686 = .000564 \] \[ x_f = \frac{.000564}{.03} = .0188 \] So, - \( C_{A2} = .075 - .0188 = .0562\,mol/L \) - \( C_{R2} = .025 + .0188 = .0438\,mol/L \) --- ## **Step 4: Final Conversion** Initial \( C_{A} = .1 \), Final \( C_{A2} = .0562 \) \[ X = \frac{C_{A} - C_{A2}}{C_{A}} = \frac{.1 - .0562}{.1} = .438 \] --- ## **Step 5: Is This Configuration Possible?** **Yes**, it is possible, **as the final conversion is less than equilibrium** (which is when \( .02C_A = .01C_R \)). The negative rate would only occur

# Step-by-Step Solution to the Liquid Phase Reaction Problem ## Given Data - **Reaction:** \( A \rightarrow R \) (series: CSTR → PFR) - **Rate Law:** \( -r_A = 0.02C_A - 0.01C_R \) - **Volumes:** \( V_{CSTR} = V_{PFR} = 200\,L \) - **Initial Concentration:** \( C_A = 0.1\,mol/L \) - **Volumetric Flow Rate:** \( Q = 100\,L/min \) ## Step 1: CSTR Material Balance For a Continuous Stirred Tank Reactor (CSTR), we can write the material balance as follows: ### Material Balance Equation \[ F_A - F_{A1} + r_A V = 0 \] Where: - \( F_A = Q \cdot C_A \) (inlet molar flow of A) - \( F_{A1} = Q \cdot C_{A1} \) (outlet molar flow of A) - \( r_A = 0.02C_{A1} - 0.01C_{R1} \) (rate at outlet) ### Substitute Values \[ 100(C_A - C_{A1}) + 200(0.02C_{A1} - 0.01C_{R1}) = 0 \] Let \( C_{R1} = C_A - C_{A1} \): \[ 100(0.1 - C_{A1}) + 200(0.02C_{A1} - 0.01(C_A - C_{A1})) = 0 \] ### Solve for \( C_{A1} \) Rearranging gives: \[ 100(0.1 - C_{A1}) + 40C_{A1} - 20(0.1 - C_{A1}) = 0 \] This simplifies to: \[ 80 - 60C_{A1} + 20C_{A1} = 0 \] \[ 80 = 40C_{A1} \] \[ C_{A1} = 2C_A \] This is not feasible since \( C_{A1} \) cannot exceed \( C_A \). ## Step 2: Check for Maximum Conversion in CSTR Using the conversion \( X \): \[ C_{A1} = C_A(1 - X_1) \] Substituting into the CSTR balance: \[ Q(C_A - C_{A1}) = r_{A1} V \] Gives: \[ 100(0.1 - C_{A(1 - X_1)}) = V(0.02C_{A(1 - X_1)} - 0.01C_{R1}) \] ## Final Conversion Calculation By substituting and rearranging, we find: \[ X = \frac{C_A - C_{A2}}{C_A} \] After solving through both CSTR and PFR, we find the final concentration: - \( C_{A2} = 0.0562\,mol/L \) - Thus, the conversion \( X \) is calculated as: \[ X = \frac{0.1 - 0.0562}{0.1} = 0.438 \] ## Step 3: Is This Configuration Feasible? **Yes**, the configuration is feasible. The final conversion is less than what would lead to reverse reactions dominating, ensuring the reaction proceeds forward. ## Conclusion The final conversion achieved in this CSTR-PFR system is approximately 43.8%, indicating the design is operationally viable.