Chapter 6: Isothermal Reactor Design: Molar Flow Rates

Gas Phase PFR

Problem:

The elementary irreversible gas phase reaction 2A -> B is carried out isothermally with no pressure drop in a PFR.

Given:

\( y_{A0} = 1 \)

\( P_0 = 8.2 \, \text{atm} \)

\( T_0 = 227^\circ C \)

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

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

Use measures other than conversion to plot molar flow rates of A and B down the reactor.

Hint 1: What are the mole balances on A and B and the rate law?

mole balances on A ?

(a) $\frac{dF_{A}}{dV} = 2r_{A}$

(b) $\frac{dF_{A}}{dV} = -2r_{A}$

(c) $\frac{dF_{A}}{dV} = -r_{A}$

(d) $\frac{dF_{A}}{dV} = r_{A}$

mole balances on B ?

(a) $\frac{dF_{B}}{dV} = r_{A}$

(b) $\frac{dF_{B}}{dV} = -2r_{A}$

(c) $\frac{dF_{B}}{dV} = \frac{-r_{A}}{2}$

(d) $\frac{dF_{B}}{dV} = \frac{r_{A}}{2}$

rate law ?

(a) $r_{A} = -k{C_{A}}^2$

(b) $r_{A} = k{C_{A}}^2$

(c) $r_{A} = kC_{A}$

Hint 2: What are the concentrations of A and B?

(a) $C_{A} = C_{T0}\frac{F_{A}}{F_{B}}p\frac{T_{0}}{T}$

(b) $C_{A} = C_{T0}\frac{F_{A}}{F_{A}+F_{B}}p\frac{T_{0}}{T}$

Hint 3: What is the rate of formation of B?

(a) $r_{B} = \frac{1}{2}C_{T0}\frac{F_{B}}{F_{T}}p\frac{T_{0}}{T}$

(b) $r_{B} = -2C_{T0}\frac{F_{B}}{F_{T}}p\frac{T_{0}}{T}$

(c) $r_{B} = -\frac{1}{2}C_{T0}\frac{F_{B}}{F_{T}}p\frac{T_{0}}{T}$

Hint 4: What are the combined mole balance, rate law and stoichiometry?

(a) $\frac{dF_{A}}{dV} = k_{A}C_{T0}\frac{F_{A}}{F_{A}+F_{B}}$

(b) $\frac{dF_{A}}{dV} = -k_{A}C_{T0}\frac{F_{A}}{F_{A}+F_{B}}$

Hint 5: What is the polymath program?

Full Solution

Hint 1 - Mole Balances and Rate Law

Mole Balances

\( \frac{dF_A}{dV} = r_A \)

\( \frac{dF_B}{dV} = r_B \)

Rate Law

\( -r_A = k_A C_A^2 \)

Back to Hints

Back to Problem

Hint 2 - Concentrations

Stoichiometry

\( \nu = \nu_0 \left( \frac{F_T}{F_{T0}} \right) \frac{P_0 T}{P T_0} \)

\( C_A = C_{T0} \frac{F_A}{F_T} \frac{P}{P_0} \frac{T}{T_0} \)

For Isothermal and No DP

\( C_A = C_{T0} \frac{F_A}{F_T} \)

\( C_B = C_{T0} \frac{F_B}{F_T} \)

\( F_T = F_A + F_B \)

Back to Hints

Back to Problem

Hint 3 - Rate of Formation of B

For \( A + \frac{b}{a} B \rightarrow \frac{c}{a} C + \frac{d}{a} D \)

\( \frac{r_A}{-a} = \frac{r_B}{-b} = \frac{r_C}{c} = \frac{r_D}{d} \)

For \( 2A \rightarrow B \)

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

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

Back to Hints

Back to Problem

Hint 4 - Combined Mole Balances, Rate Law and Stoichiometry

Combine

\( \frac{dF_A}{dV} = -k_A C_A^2 = -k_A C_T^2 \left( \frac{F_A}{F_T} \right)^2 \)

\( \frac{dF_B}{dV} = \frac{1}{2} k_A C_A^2 = \frac{k_A}{2} C_T^2 \left( \frac{F_A}{F_T} \right)^2 \)

\( F_T = F_A + F_B \)

\( C_T^0 = \left( \frac{8.4 \, \text{atm}}{0.082 \, \text{atm dm}^{-3} \, \text{mol}^{-1} \, \text{k}^{-1}} \right) \cdot 500 \, \text{K} = 0.2 \, \text{mol/dm}^3 \)

Back to Hints

Back to Problem

Hint 5 - Polymath Code

[Note: To=227C=500K]

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

\( V = 0 \quad F_A = F_{A0} \quad F_B = 0 \)

\( V_f = 100 \, \text{dm}^3 \)

Use Polymath to solve

One can back calculate X

\( X = \frac{F_{A0} - F_A}{F_{A0}} \)

See the Polymath Solution Below

Back to Hints

Back to Problem

Full Solution

Mole Balances

\( \frac{dF_A}{dV} = r_A \)

\( \frac{dF_B}{dV} = r_B \)

Rate Law

\( -r_A = k_A C_A^2 \)

Stoichiometry

\( \nu = \nu_0 \left( \frac{F_T}{F_{T0}} \right) \frac{P_0 T}{P T_0} \)

\( C_A = C_{T0} \frac{F_A P}{F_T P_0 T_0 T} \)

For \(A + \frac{b}{a} B \rightarrow \frac{c}{a} C + \frac{d}{a} D\)

\( -r_A = -r_B = r_C = r_D \)

\( \frac{-r_A}{a} = \frac{-r_B}{b} = \frac{r_C}{c} = \frac{r_D}{d} \)

For 2A->B

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

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

For Isothermal and No DP

\( C_A = C_{T0} \frac{F_A}{F_T} \)

\( C_B = C_{T0} \frac{F_B}{F_T} \)

\( F_T = F_A + F_B \)

Combine

\( \frac{dF_A}{dV} = -k_A C_A^2 = -k_A C_{T0} \left( \frac{F_A}{F_T} \right)^2 \)

\( \frac{dF_B}{dV} = \frac{1}{2} k_A C_A^2 = \frac{k_A}{2} C_{T0} \left( \frac{F_A}{F_T} \right)^2 \)

\( F_T = F_A + F_B \)

\( C_{T0} = \left( \frac{8.4 \, \text{atm}}{0.082 \, \text{atm} \, \text{dm}^3 \, \text{mol} \cdot \text{k}} \right) \frac{1}{500 \, \text{K}} = 0.2 \, \text{mol/dm}^3 \)

[Note: \( T_0 = 227^\circ \text{C} = 500K \)]

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

\( V = 0 \)

\( F_A = F_{A0} \)

\( F_B = 0 \)

\( V_f = 100 \, \text{dm}^3 \)

Use Polymath to solve

One can back calculate X

\( X = \frac{F_{A0} - F_A}{F_{A0}} \)

See the Polymath Solution Below

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