Chapter 6: Isothermal Reactor Design: Molar Flow Rates

Liquid Phase CSTR

Chemical reactor diagram showing a reaction 2A → B in a cylindrical reactor. The inputs are C_A0 = 0.02 mol/dm^3 and v_0 = 25 dm^3/s, with a reactor volume V = 500 dm^3. The reaction rate constant is k_A = 10 dm^3/(mol·s). The output concentration C_A is to be determined.

\( C_{A0} = 0.02 \, \frac{\text{mol}}{\text{dm}^3} \)

\( v_0 = 25 \, \frac{\text{dm}^3}{\text{s}} \)

\( 2A \to B \)

\( V = 500 \, \text{dm}^3 \)

\( C_A = ? \)

\( k_A = 10 \, \frac{\text{dm}^3}{\text{mol} \cdot \text{s}} \)

Hint 1: What are the mole balances for A and B?

Hint 2: What is the rate law?

$ (a) V = \frac{F_{A0} - F_{A}}{-r_{A}} \textrm{,} V = \frac{F_{B}}{r_{B}}$

$ (b) V = \frac{F_{A0} - F_{A}}{r_{B}} \textrm{,} V = \frac{F_{B}}{-r_{B}}$

$ (c) V = \frac{\upsilon_{0}(C_{A0} - C_{A})}{-r_{A}} \textrm{,} V = \frac{\upsilon_{0}C_{B}}{-r_{A}}$

Hint 3: What is the combined rate law and mole balance?

$ (a) V = \frac{\upsilon_{0}(C_{A0}-C_{A})}{k{C_{A}}^2}$

$ (b) V = \frac{\upsilon_{0}C_{B}}{k{C_{A}}^2} $

Hint 1

Mole Balances

\( V = \frac{F_{A0} - F_A}{r_A} \)

\( V = \frac{F_{B0} - F_B}{-r_B} \)

Back to Problem

Hint 2

Rate Law

\(-r_A = k_A C_A^2\)

Stoichiometry Liquid n = n_o

\(F_A = C_A \nu_o\)

\(F_{A0} = C_{A0} \nu_o\)

\(\frac{-r_A}{2} = \frac{r_B}{1}\)

\(r_B = \frac{1}{2} (-r_A) = \frac{k_A}{2} C_A^2\)

Back to Problem

Hint 3

Combine

\(-r_A = k_A C_A^2\)

\(V = \nu_o \left[ \frac{C_{A0} - C_A}{k_A C_A^2} \right]\)

Back to Problem

Solution

Mole Balances

\( V = \frac{F_{A0} - F_A}{r_A} \)

\( V = \frac{F_{B0} - F_B}{-r_B} \)

Rate Law

\(-r_A = k_A C_A^2\)

Stoichiometry Liquid n = n_o

\(F_A = C_A \nu_o\)

\(F_{A0} = C_{A0} \nu_o\)

\(\frac{-r_A}{2} = \frac{r_B}{1}\)

\(r_B = \frac{1}{2} (-r_A) = \frac{k_A}{2} C_A^2\)

Combine

\(-r_A = k_A C_A^2\)

\(V = \nu_o \left[ \frac{C_{A0} - C_A}{k_A C_A^2} \right]\)

Dividing by v_o, and letting t = \( \frac{V}{\nu_0} \), the

\( r = \frac{C_{A0} - C_A}{k_A C_A^2} \)

Rearranging

\( r k_A C_A^2 + C_A - C_A0 = 0 \)

\( C_A = \frac{-1 + \sqrt{1 + 4r k_A C_A0}}{2r k_A} \)

\( r = \frac{500 \, \text{dm}^3}{25 \, \text{dm}^3/\text{s}} = 20 \, \text{s} \)

\( r k = 20 \, \text{s} \times 10 \, \frac{\text{dm}^3}{\text{mol} \cdot \text{s}} = 200 \, \frac{\text{dm}^3}{\text{mol}} \)

\( r k C_{A0} = 20 \, \text{s} \times 10 \, \frac{\text{dm}^3}{\text{mol} \cdot \text{s}} \times 0.2 \, \frac{\text{mol}}{\text{dm}^3} = 40 \)

\( C_A = \frac{-1 + \sqrt{1 + 4 \cdot 40}}{2 \cdot 200} = 0.029 \, \frac{\text{mol}}{\text{dm}^3} \)

Back calculate X

\( X = \frac{C_{A0} - C_A}{C_{A0}} = \frac{0.2 - 0.029}{0.2} = 0.855 \)

Back to Chapter 6