Chapter 8: Multiple Reactions
What is wrong with this solution?
Problem
Consider the elementary gas phase reactions
\( A \overset{k_{1A}}{\underset{k_{2A}}{\rightleftharpoons}} 2B \xrightarrow{k_{3B}} 3C \)
\( 2B \xrightarrow{k_{4B}} D \)
Set up the equations to calculate the concentration of each species as a function of time in a constant volume batch reactor. The reaction is carried out isothermally.
Solution
A combined mole balance(1) and rate laws(2)
\( \frac{dN_A}{dt} = -k_{1A} \left( C_A - \frac{C_B^2}{K_{e1}} \right) \)
\( \frac{dN_B}{dt} = 2k_{1A} \left( C_A - \frac{C_B^2}{K_{e1}} \right) - \frac{k_{3B}}{3} C_B^2 - \frac{k_{4B}}{2} C_B^2 \)
\( \frac{dN_C}{dt} = +\frac{2}{3} k_{3B} C_B^2 \)
(3) Stoichiometry
\( \frac{dN_D}{dt} = 2k_{4B} C_B^2 \)
\( C_A = C_{T0} \frac{N_A}{N_T} \frac{P}{P_0} \)
\( C_B = C_{T0} \frac{N_B}{N_T} \frac{P}{P_0} \)
\( N_T = N_A + N_B + N_C + N_D \)
\( \frac{P}{P_0} = \frac{N_T}{N_{T0}} = \frac{N_T}{N_{A0} + N_{B0}} \)
at \( t = 0 \) then \( N_A = N_{A0} \) and \( N_{B0} = N_{C0} = N_{D0} = 0 \)
These equations will be typed into Polymath.
What five things are wrong with each of the first three steps of the multiple reaction algorithm?
Answer
(1)Mole Balance: V missing on right hand side of each of the combined mole balance and rate law equations (1) through (4).
(2)Stoiciometry: The gas reaction occurs in a constant volume rigid container, V = V0. Therefore the equations [5) and (6)] for the concentration of the reacting species are not correct
\( C_A = \frac{N_A}{V} = \frac{N_A}{V_0} \)
(3)Rate laws, relative rates, and net rates: The net rate of reaction is not correct for B,C, and D, i.e., equations (2), (3), and (4).
Correct Solutions
Since V = V0
\( \frac{1}{V_0} \frac{dN_A}{dt} = \frac{dC_A}{dt} \), etc.
(1) \( A \rightarrow 2B \quad -r_{1A} = k_{1A} C_A \)
(2) \( 2B \rightarrow A \quad -r_{2A} = k_{2A} C_B^2 \)
(3) \( 2B \rightarrow 3C \quad -r_{3B} = k_{3B} C_B^2 \)
(4) \( 2B \rightarrow D \quad -r_{4B} = k_{4B} C_B^2 \)
Mole Balance
(1) \( \frac{dC_A}{dt} = r_A \)
(2) \( \frac{dC_B}{dt} = r_B \)
(3) \( \frac{dC_C}{dt} = r_C \)
(4) \( \frac{dC_D}{dt} = r_D \)
Net Rates and Rate Laws
Net Rates
Species A
\( r_A = r_{1A} + r_{-1A} = -k_{1A} \left[ C_A - \frac{C_B^2}{K_{1e}} \right] \)
\( r_B = r_{1B} + r_{-1A} + r_{2B} + r_{3B} \)
Species B
\( \frac{r_{1B}}{2} = r_{1A}, \quad r_{1B} = 2k_{1A} C_A \)
\( \frac{r_{2B}}{2} = r_{2A}, \quad r_{2B} = 2k_{2A} C_B^2 \)
\( r_{3B} = -k_{3B} C_B^2 \)
\( r_{4B} = -k_{4B} C_B^2 \)
\( r_B = 2 \left[ k_{1A} C_A - k_{2A} C_B^2 \right] - k_{3B} C_B^2 - k_{4B} C_B^2 \)
\[ \boxed{ r_B = k_{1A} \left[ C_A - \frac{C_B^2}{K_{1e}} \right] - k_{3B} C_B^2 - k_{4B} C_B^2 } \]
Species C
\( r_C = r_{3C} \)
\( \frac{r_{3C}}{3} = \frac{-r_{3B}}{2} \)
\[ \boxed{ r_{3C} = \frac{3}{2} k_{3B} C_B^2 } \]
Species D
\( r_D = r_{4D} \)
\( \frac{r_{4D}}{1} = \frac{-r_{4B}}{2} \)
\[ \boxed{ r_{4D} = \frac{1}{2} k_{4D} C_B^2 } \]
Summary
\( \frac{dC_A}{dt} = -k_{1A} \left[ C_A - \frac{C_B^2}{K_{1e}} \right] \)
\( \frac{dC_B}{dt} = 2k_{1A} \left[ C_A - \frac{C_B^2}{K_{1e}} \right] - k_{3B} C_B^2 - k_{4B} C_B^2 \)
\( \frac{dC_C}{dt} = \frac{3}{2} k_{3B} C_B^2 \)
\( \frac{dC_D}{dt} = \frac{1}{2} k_{4B} C_B^2 \)
Stoichiometry
The pressure inside the vessel is
\( \frac{P}{P_0} = \frac{N_T}{N_{T0}} = \frac{V C_T}{V C_{T0}} = \frac{C_A + C_B + C_C + C_D}{C_{A0}} \)