Chemical Reaction Engineering
With multiple reactions, either molar flow or number of moles must be used in setting up the balance equations (no conversion!)
Mole balance: Write balance for each species i = 1 \ to \ N
\frac{dF_i}{dW} = r'_i
Rate law
-r'_A = kC_A^\alpha C_B^\beta; \qquad \frac{r'_A}{-a} = \frac{r'_B}{-b} = \frac{r'_C}{c} = \frac{r'_D}{d}
Stoichiometry C_i = \frac{C_{A0} (\Theta_i + \nu_i X)} {(1 + \epsilon X)} \left( \frac{P}{P_0} \right) \left( \frac{T_0}{T} \right)
Pressure: \frac{dP}{dW} = -\frac{\alpha}{2p} \left( \frac{T}{T_0} \right) \frac{F_T}{F_{T0}}
Total molar flow rate: F_T = \sum_{i=1}^N F_i
Combine:
Collate all equations from steps 1 to 3 to yield a system of equations
Evaluate:
Use ODE solver to solve the system of equations obtained in step 4.
Competing reactions
Reactant is consumed by two different pathways to form different products
\ce{A ->[k_1] D} \\ \ce{A ->[k_2] U}
Examples
Ethylene oxidation
\begin{align*} \ce{C2H4 + O2} & \ce{-> C2H4O} && \text{ethylene oxide} \\ \\ \ce{C2H4 + O2 } & \ce{-> 2 CO2 + 2 H2O} && \text{complete combustion} \end{align*}
Fischer Tropsch synthesis
\begin{align*} \ce{CO + 3 H2 } & \ce{->[k_1] CH4 + H2O} \\ \\ \ce{CO + 2 H2 } & \ce{->[k_2] [C_nH_{2n}]_n + H2O} \end{align*}
Consecutive reactions
Reactions where reactant forms an intermediate product, which reacts further to form another product.
\ce{A ->[k_1] D ->[k_2] U} \qquad
Example
\begin{align*} \ce{C2H4O + NH3} & \ce{-> (HOCH2CH2)NH2} && \text{(mono ethanolamine)}\\ \\ \ce{(HOCH2CH2)NH2 + C2H4O} & \ce{->} \underset{\textcolor{RoyalBlue}{\text{desired product}}}{\ce{(HOCH2CH2)2NH}} && \text{(di ethanolamine)} \\ \\ \ce{(HOCH2CH2)2NH + C2H4O} & \ce{-> (HOCH2CH2)3N} && \text{(tri ethanolamine)} \end{align*}
Reactions that occur at the same time
Neither the products nor the reactants react with themselves or one another
\ce{A ->[k_1] D} \\ \ce{C ->[k_2] U}
Example: Cracking of crude oil
Hundreds of reactions
\begin{align*} \ce{C15H32} & \ce{-> C12H26 + C3H6 }\\ \\ \ce{C8H18} & \ce{-> C6H14 + C2H2 } \end{align*}
Multiple reactions involving a combination of series, parallel, and/or independent reactions
\begin{align*} \ce{A + B} & \ce{->[k_1] C + D} \\ \ce{A + C } & \ce{->[k_2] E} \\ \ce{E } & \ce{->[k_3] G} \end{align*}
Example: Formation of butadiene from ethanol
\begin{align*} \ce{C2H5OH} & \ce{-> C2H4 + H2O }\\ \\ \ce{C2H5OH} & \ce{-> CH3CHO + H2 }\\ \\ \ce{C2H4 + CH3CHO} & \ce{-> C4H6 + H2O } \end{align*}
Parallel reactions \begin{align*} \ce{A} & \ce{-> D } && \leftarrow \text{desired product}\\ \\ \ce{A} & \ce{-> u } && \leftarrow \text{undesired byproduct}\\ \end{align*}
Series reactions \ce{A ->} \underset{\text{desired product}}{\ce{B}} \ce{-> U}
Minimize formation of U and maximize formation of D
Greater the amount of U \leftarrow lower production of desired product, higher cost of separation \leftarrow lower profits
S_{D/U} = \frac{r_D}{r_U} = \frac{ \text{rate of formation of D}}{ \text{rate of formation of U}}
\bar{S}_{D/U} = \frac{F_D}{F_U} = \frac{ \text{Exit molar flow rate of D}}{ \text{Exit molar flow rate of U}}
For CSTR:
F_D = r_D V
F_U = r_U V
\therefore F_D/F_U = r_D/r_U = S_{D/U} = \bar{S}_{D/U}
For batch reactor:
\bar{S}_{D/U} = N_D/N_U
N_D, N_U: Number of moles of D and U at the end of the reaction.
Ratio of the reaction rate of a given product to the reaction rate of key reactant A
Y_D = \frac{r_D}{-r_A} = \frac{ \text{Rate of formation of D}}{ \text{Rate of consumption of A}}
Ratio of moles of product formed at the end of the reaction to the number of moles of the key reactant A, that have been consumed.
\bar{Y}_{D} = \underset{\text{For CSTR}}{\frac{F_D}{F_{A0} - F_A}} = \underset{\text{For batch reactor}}{\frac{N_D}{N_{A0} - N_A}}
Gives insight into problem
Often conflicts with selelctivity
Ideal world \Rightarrow make as much D as possible simultaneously minimize U
Practical experiece \Rightarrow Greater the conversion, more the undesired product
Not used in solving multiple reaction problems, but calculated later for analysis
For species A X_A = \underset{\text{Flow system}}{\frac{F_{A0} - F_A}{F_{A0}}} \qquad X_A = \underset{\text{Batch system}}{\frac{N_{A0} - N_A}{N_{A0}}}
For species B X_B = \underset{\text{Flow system}}{\frac{F_{B0} - F_B}{F_{B0}}} \qquad X_B = \underset{\text{Batch system}}{\frac{N_{B0} - N_B}{N_{B0}}}
Derive expression for conversion for a semibatch system where B is fed to A
Number each and every reaction separately.
Mole balance on each and every species.
F_{j0} - F_j + \int^V r_j dV = \frac{dN_j}{dt}
For every reaction write rate law. Calculate the net rate of reaction and relative rates
For component j: r_j = \sum_{i=1}^N r_{i,j}
Stoichiometry
Combine:
Collate all equations from steps 2 to 4 to yield a system of equations
Evaluate:
Use ODE solver to solve the system of equations obtained in step 4.
Can be applied to parallel, series, independent, and complex reactions.
For liquid systems concentration is usually preferred variable for mole balance.
Consider two competing reactions
\begin{align*} \ce{A -> D} & \text{(desired)} \qquad && r_D = k_D C_A^{\alpha_1}\\ \\ \ce{A -> U} & \text{(undesired)} \qquad && r_U = k_U C_A^{\alpha_2} \end{align*}
Net rate of disappearance of A
-r_A = r_D + r_U= k_D C_A^{\alpha_1} + k_U C_A^{\alpha_2}
S_{D/U} = \frac{r_D}{r_U} = \frac{k_D}{k_U} C_A^{\alpha_1 - \alpha_2}
\alpha_1, \alpha_2 are positive orders
\alpha_1 - \alpha_2 = a
S_{D/U} = \frac{k_D}{k_U} C_A^{a}
To maximize selectivity we wan to carry out the reaction in a manner that will keep C_A as high as possible during the reaction.
Use PFR or batch reactor
Gas phase: Use high pressure, run without inerts
Liquid phase: minimize diluent
\alpha_2 - \alpha_1 = b
S_{D/U} = \frac{k_D}{k_U} \frac{1}{C_A^{b}}
To maximize selectivity we wan to carry out the reaction in a manner that will keep C_A as low as possible during the reaction.
Use CSTR or dilute feed stream
Recycle reactor
Need information on activation energy
Sensitivity to temperature for fixed concentrations (C_A^a is constant)
S_{D/U} \approx \frac{k_D}{k_U} = \frac{A_D}{A_U} e^{-[(E_D - E_U)/RT]}
E_D > E_U
E_D < E_U
k_U increases more rapidly than k_D with increase in temperature
Reaction should be carried out at lower temperature to maximize S_{D/U}
The temperature should not be very low as it might affect reaction extent. Reaction may not proceed at low temperature.
Consider two competing parallel reactions
\begin{align*} \ce{A + B -> D} & \text{(desired)} \qquad && r_D = k_D C_A^{\alpha_1} C_B^{\beta_1}\\ \\ \ce{A + B -> U} & \text{(undesired)} \qquad && r_U = k_U C_A^{\alpha_2} C_B^{\beta_2} \end{align*}
Net rate of disappearance of A
-r_A = r_D + r_U= k_D C_A^{\alpha_1} C_B^{\beta_1} + k_U C_A^{\alpha_2} C_B^{\beta_2}
Selectivity
S_{D/U} = \frac{r_D}{r_U} = \frac{k_D}{k_U} C_A^{\alpha_1 - \alpha_2} C_B^{\beta_1 - \beta_2}
\alpha_1 - \alpha_2 = a; \beta_1 - \beta_2 = b
S_{D/U} \propto C_A^{a} C_B^{b}
Use high C_A, C_B
Configurations:
S_{D/U} \propto \frac{C_A^{a}}{C_B^{b}}
Use high C_A, low C_B
Configurations:
\alpha_1 - \alpha_2 = a; \beta_1 - \beta_2 = b
S_{D/U} \propto C_B^{b} C_A^{a}
Use high C_B, low C_A
Configurations:
S_{D/U} \propto \frac{1}{C_A^{a} C_B^{b}}
Use low C_A, C_B
Configurations:
CSTR
Feed diluted streams with inerts
Low pressure (gas phase)
Tubular reactor with large recycle ratio
Can be used for highly exothermic reactions. The recycle stream is cooled and returned to the reactor to dilute and cool inlet stream.
Such configuration helps in avoiding hotspots and runaway reactions.
The most important variable is time
Consider: \ce{A ->[k_1] B ->[k_2] C}.
B is the desired product
If k_1 \ll k_2: First reaction is slow
If k_1 \gg k_2: First reaction is fast
If reaction is allowed to proceed for a long time, desired product B will be converted to undesired product C.
Accuracy of prediction for time required to carry out the reaction is vital.
Consider the reaction
\ce{A ->[k_1] B ->[k_2] C}.
B is the desired product, C is waste product.
We are interested in
Number of reactions
The series reaction can be written as two reactions
\ce{A -> B}; -r_{1A} = k_1 C_A.
\ce{B -> C}; -r_{2B} = k_2 C_B.
Mole balance \frac{dN_i}{dt} = r_i V
For constant volume batch reactor
\frac{dC_A}{dt} = r_A = -k_1 C_A \tag{1}
\frac{dC_B}{dt} = r_B = k_1 C_A - k_2 C_B \tag{2}
\frac{dC_C}{dt} = r_C = k_2 C_C \tag{3}
Chemical Reaction Engineering