Introduction to Chemical Reaction Kinetics#

Introduction#

The kinetics of a chemical reaction describes the rate at which the reaction proceeds as a function of the conditions in the reactor vessel. Kinetic rate laws have units of \(\star\)mol/volume-time, and their mathematical form varies significantly with the type of reaction being considered.

In this lecture, we will:

Concentration balance equations#

When describing systems with chemical reactions where we write reaction rate expressions in terms of concentration, e.g., mole per unit volume basis, it is inconvient to use mole or mass based units. In these cases, we need a new type of balance equation, the concentration balance.

Starting from species mole balance equations, we formulate the concentration balance by replacing moles with concentration units (Definition 29):

Definition 29 (Open species concentration balance)

Let \(n_{i}\) denote the number of moles of chemical component \(i\) in the system (units: \(\star\)moles, e.g., mmol or \(\mu\)mol). Further, denote the chemical species in the system with the set \(\mathcal{M}\), and the set of streams flowing into (or from) the system as \(\mathcal{S}\).

Then, the number of moles of chemical component \(i\) in the system is described by an open species mass balance equation:

\[\sum_{s\in\mathcal{S}}\nu_{s}\dot{n}_{s,i} + \dot{n}_{G,i} = \frac{dn_{i}}{dt} \qquad\forall{i}\in\mathcal{M}\]

However, we can re-write the number of moles of species \(i\) as:

\[n_{i} = C_{i}V\qquad\forall{i}\in\mathcal{M}\]

where \(C_{i}\) denotes the concentration of species \(i\) (units: mole per volume), and \(V\) (units: volume) denotes the volume of the system. Thus, we can re-write the species mole balance in concentration units as:

(69)#\[\sum_{s\in\mathcal{S}}\nu_{s}C_{s,i}\dot{V}_{s} + \dot{C}_{G,i}V = \frac{d}{dt}\left(C_{i}V\right)\qquad\forall{i}\in\mathcal{M}\]

where \(\dot{V}_{s}\) denotes the volumetric flow rate for stream \(s\) (units: volume/time), \(C_{s,i}\) denotes the concentration of component \(i\) in stream \(s\) (units: concentration), and \(\dot{C}_{G,i}\) denotes the rate of generation of component \(i\) by chemical reaction (units: concentration/time).

The concentration generation terms in Eqn. (69) describe the impact of chemical reactions. Previously, we described reactions using the open extent of reaction \(\dot{\epsilon}_{i}\) to describe chemical reactions. However, we’ve never defined what \(\dot{\epsilon}_{i}\).

Open extent and the reaction rate#

The open extent of reaction \(\dot{\epsilon}_{i}\) has two interpretations, a steady-state thermodynamic interpretation, and a time-dependent kinetic one in which \(\dot{\epsilon}_{i}\) is related to the rate of reaction; let’s start with the kinetic interpretation.

Definition 30 (Concentration generation terms)

Let the set of species \(\mathcal{M}\) participate in the chemical reaction set \(\mathcal{R}\). Further, define the extent of reaction for reaction \(j\) as:

(70)#\[\epsilon_{j}\equiv\hat{r}_{j}V\qquad\forall{j}\in\mathcal{R}\]

where \(\hat{r}_{j}\) denotes the reaction rate for reaction \(j\) (units: mole/volume-time). Then, the generation terms for species \(i\) in the concentration balance equation can be written as:

(71)#\[\dot{C}_{G,i}V = \sum_{j\in\mathcal{R}}\sigma_{ij}\hat{r}_{j}V\qquad\forall{i}\in\mathcal{M}\]

where \(\hat{r}_{j}\) denotes the rate of reaction \(j\) (units: mol/volume-time) and \(\sigma_{ij}\) denotes the stoichiometric coefficient for species \(i\) in reaction \(j\):

  • If \(\sigma_{ij}>0\) then species \(i\) is produced by reaction \(j\), i.e., species \(i\) is a product of reaction \(j\)

  • If \(\sigma_{ij}=0\) then species \(i\) is not connected to reaction \(j\)

  • If \(\sigma_{ij}<0\) then species \(i\) is consumed by reaction \(j\), i.e., species \(i\) is a reactant of reaction \(j\).

General model of chemical reaction kinetics#

The rate of a chemical reaction can be dependent upon the concentration of both the reactants, and potnetially the products as well as constants called rate constants. Thus, a general model for the net rate of reaction, involving the species set \(\mathcal{M}\), is given by:

(72)#\[\hat{r}_{1} = k_{1}\prod_{i\in\mathcal{M}}\left[X_{i}\right]^{\alpha_{i1}} - k_{1}^{\prime}\prod_{j\in\mathcal{M}}\left[X_{j}\right]^{\beta_{j1}}\]

The first product term in Eqn. (72) describes the forward rate (reactants \(\rightarrow\) products), while second product term decribes the reverse rate (prodcuts \(\rightarrow\) reactants). The terms \(\left[X_{i}\right]\) denotes the concentration of component \(i\in\mathcal{M}\), and \(\alpha_{i1}\) and \(\beta_{i1}\) denote the reaction order of component \(i\) in reaction \(\hat{r}_{1}\). The values of the reation orders \(\alpha_{i1}\) and \(\beta_{i1}\) depend upon the reaction, and may not always be integer values. Finally, the quantities \(k_{1}\) and \(k_{1}^{\prime}\) denote the rate constant for the forward and reverse directions, respectively. Rate constants have various units depending upon the values of \(\alpha_{i1}\) and \(\beta_{i1}\).

Remark 7 (Are reaction orders integers?)

There is no requirement that \(\alpha_{i1}\) and \(\beta_{i1}\) be integers (or the stoichiometric coefficients). This is often only true when we have a complete (or simplified) picture of the chemistry that is occurring. In real-life, fractional values for \(\alpha_{i1}\) and \(\beta_{i1}\) are common in many application areas.

Mass action kinetics#

Let’s consider a simplification to the general rate law in Eqn. (72). Assume the forward and reverse reactions are elementary reactions; an elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step where no reaction intermediates are required to describe the reaction on a molecular scale.

Elementary reactions can be described by the law of mass action. The law of mass action assumes that the net rate of a chemical reaction is proportional to the concentration of the components participating in the reaction raised to the \(-{1}\times\) the stoichiometric coefficient of that component in a particular reaction direction:

Definition 31 (Mass action kinetics)

Let the set of species \(\mathcal{M}\) participate in a reversible elementary reaction with the rate \(\hat{r}_{j}\). Further, let the reactants of \(\hat{r}_{j}\) be in set \(\mathcal{M}^{-}\) and the products of \(\hat{r}_{j}\) be in set \(\mathcal{M}^{+}\), where \(\mathcal{M}^{-}\cup\mathcal{M}^{+} = \mathcal{M}\).

Then, the law of mass action for reaction \(j\) (that could involve autocatalysis) is given by:

(73)#\[\hat{r}_{j} = k_{j}\prod_{i\in\mathcal{M}^{-}}\left[X_{i}\right]^{-\sigma_{ij}} - k^{\prime}_{j}\prod_{k\in\mathcal{M}^{+}}\left[X_{k}\right]^{\sigma_{kj}}\]

where \(k_{j}\) and \(k^{\prime}_{j}\) are the forward and reverse rate constants, \(\left[X_{i}\right]\) is the concentration of species \(i\), and \(\sigma_{kj}\) is the stoichiometric coefficient governing the species \(k\) in reaction \(j\).

Let’s consider an example of a simple mass action reaction in a well-mixed closed system (Example 10):

Enzyme catalyzed reactions#

Enzymes are biological polymers, largely composed of the 20 naturally occurring amino acid building, that catalyze biochemical reactions. There are seven classes of enzymes, and each enzyme class can carry out a specific type of chemistry:

Name

Reaction

Description

Oxidoreductases

\(A_{o}\) + \(B_{r}\) \(\rightarrow\) \(A_{r}\) + \(B_{o}\)

Catalyze redox reactions

Transferases

A-B + C \(\rightarrow\) A + C-B

Catalyze the transfer of groups between substrates

Hydrolases

A-B + H2O \(\rightarrow\) A-H + B-OH

Catalyze the hydrolysis of substrates

Lyases

A-B \(\rightarrow\) A + B

Catalyzes the removal of a group from a substrate

Isomerases

A-B-C \(\rightarrow\) A-C-B

Catalyzes the formation of isomers

Ligases

A + B + ATP \(\rightarrow\) A-B + ADP + Pi

Catalyze the joining of two substrates

Translocases

A-out \(\rightarrow\) A-in

Catalyze the movement of ions or molecules across membranes

In general, the kinetics of enzyme-catalyzed reactions is complex. However, there is a simple theoretical model, the Michaelis–Menten kinetic model, that is a helpful starting point for broadly understanding the factors that control the rate of an enzyme-catalyzed reaction.

Michaelis–Menten kinetics#

Let’s assume we have a well-mixed test tube containing an enzyme \(E\) (a protein that catalyzes chemical reactions), which converts substrate \(S\) (the starting compound) into product \(P\) according to three elementary reactions:

\[\begin{split}\begin{eqnarray} E+S&\rightleftharpoons&{E:S}\\ {E:S}&\longrightarrow&E+P \end{eqnarray}\end{split}\]

where enzyme must obey the relationship:

\[\left[E_{T}\right] = \left[E\right] + \left[E:S\right]\]

The term \(\left[E_{T}\right]\) denotes the total enzyme concentration in the tube, \(\left[E\right]\) denotes the free enzyme concentration (not bound to substrate) while \(\left[E:S\right]\) denotes the enzyme-substrate complex.

We can describe the binding, unbinding and catalysis steps as three elementary reactions:

Observation 26 (Michaelis–Menten elementary steps.)

The kinetics of each elementary step can be written using mass-action kinetics, i.e.,

\[\begin{split}\begin{eqnarray} r_{1} & = & k_{1}\left[E\right]\left[S\right]\\ r_{2} & = & k_{2}\left[E:S\right]\\ r_{3} & = & k_{3}\left[E:S\right] \end{eqnarray}\end{split}\]

where \(\left[\cdot\right]\) denotes a species concentration, and \(k_{j}\) denotes the rate constant governing the jth elementary reaction:

  • The elementary rate \(r_{1}\) describes the association rate between the enzyme and substrate

  • The elementary rate \(r_{2}\) represents the rate of dissociation of the enzyme-substrate complex

  • The elementary \(r_{3}\) denotes the rate of chemical conversion of the bound substrate into the product (we assume the product dissociation from the enzyme is fast).

Derivation#

To estimate the overall rate of reaction \(v\), we propose a single rate-limiting step out of the set of elementary reactions. Let’s assume that the chemical conversion rate \(r_{3}\) is the slowest step, i.e., the substrate bounces on/off the enzyme quickly, with only a tiny fraction of these binding events resulting in a successful chemical transformation. Thus, the overall rate is then given by:

\[v = k_{3}\left[E:S\right]\]

Let’s also assume that we already know (or can estimate) the rate constants \(k_{1},k_{2}\) and \(k_{3}\). When this is true, the only unknown is \(\left[E:S\right]\). However, we can relate \(\left[E:S\right]\) to variables we know (\(E_{T}\) and at least initially \(S\)) through the enzyme balance, and the pseudo-steady-state assumption for the reaction intermediate \(\left[E:S\right]\):

\[\frac{d\left[E:S\right]}{dt} = k_{1}\left[E\right]\left[S\right] - k_{2}\left[E:S\right] - k_{3}\left[E:S\right]\simeq{0}\]

Rearranging and solving for \(\left[E:S\right]\) gives the relationship:

\[\left[E:S\right]\simeq\frac{k_{1}}{k_{2}+k_{3}}\left[E\right]\left[S\right]\]

where the ratio of constants is defined as the Michaelis-Menten saturation coefficient or \(K_{M}\):

\[\frac{1}{K_{M}}\equiv\frac{k_{1}}{k_{2}+k_{3}}\]

Substituting the definition of \(K_{M}\) into the overall rate yields:

\[v = k_{3}\frac{\left[E\right]\left[S\right]}{K_{M}}\]

However, we do not know the free enzyme concentration of \(\left[E\right]\); to find \(\left[E\right]\) we subsitute \(\left[E:S\right]\) into the enzyme balance and solving for \(\left[E\right]\):

\[\left[E\right] = \frac{\left[E_T\right]K_{M}}{K_{M}+\left[S\right]}\]

Lastly, we substitute \(\left[E\right]\) into the overall rate to arrive at the final expression for \(v\):

\[v = V_{max}\frac{\left[S\right]}{K_{M}+\left[S\right]}\]

where \(V_{max}\equiv{k_{3}}\left[E_{T}\right]\).

Definition 32 (Michaelis–Menten Kinetics)

Suppose we have an enzyme \(E\) that converts substrate \(S\) (the starting compound) into product \(P\) in a well-mixed reaction. Then, the overall reaction rate for the conversion of substrate \(S\) to product \(P\) is given by:

(74)#\[v = V_{max}\frac{\left[S\right]}{K_{M}+\left[S\right]}\]

where \(K_{M}\) is the saturation constant for enzyme \(E\) and substrate \(S\), while \(V_{max}\) is the maximum rate of reaction:

(75)#\[V_{max}\equiv{k_{3}}\left[E_{T}\right]\]

where \(k_{3}\) is called the turn-over number of the enzyme \(E\) (units: 1/time) and \(\left[E_{T}\right]\) denotes the concentration of enzyme \(E\).

Limiting cases:

  • If \(S\gg{K}_{M}\), the rate \(v\simeq{V_{max}}\).

  • If \(S\ll{K}_{M}\), the rate \(v\) appears to be linear with respect to substrate concentration.

  • If \(K_{M}\simeq S\), the rate \(v\simeq 1/2~V_{max}\).

Example#

Well-mixed assumption#

Now that we have introduced kinetic expressions, let’s introduce one final key concept in this section, the Well Mixed Assumption (WMA). To understand the benefit (and cost) of the well mixed assumption, let’s manupilate the species concentration balance equations given in Definition 29.

Constant volume, steady-state concentration balances#

Suppose we have a reaction set \(\mathcal{R}\) involving the species set \(\mathcal{M}\) that is occurring in a well-mixed chemical reactor with stream set \(\mathcal{S}\). Then the concentration of component \(i\in\mathcal{M}\) is given by:

\[\frac{d}{dt}\left(C_{i}V\right) = \sum_{s\in\mathcal{S}}v_{s}C_{s,i}\dot{V}_{s} + \sum_{r\in\mathcal{R}}\sigma_{ir}\hat{r}_{r}V\qquad\forall{i}\in\mathcal{M}\]

where \(C_{s,i}\) denotes the concentration of component \(i\in\mathcal{M}\) in stream \(s\in\mathcal{S}\), \(\sigma_{ir}\) denotes the stoichiometric coefficient for component \(i\in\mathcal{M}\) in reaction \(r\in\mathcal{R}\), \(\dot{V}_{s}\) denotes the volumetric flow rate of stream \(s\in\mathcal{S}\), \(V\) denotes the volume of the reaction mixture in the reactor unit, \(v_{s}\) denotes the direction parameter for stream \(s\in\mathcal{S}\) and the quantity \(C_{i}\) denotes the concentration of component \(i\in\mathcal{M}\) in the reaction vessel. Finally, the terms \(\hat{r}_{r}\) denote the reaction rate per unit volume for reaction \(r\in\mathcal{R}\) that is occurring in the reaction vessel.

When the reaction is at constant volume, we can pull the volume \(V\) out of the accumulation term and divide by the volume to give:

\[\frac{dC_{i}}{dt} = \sum_{s\in\mathcal{S}}v_{s}C_{s,i}D_{s} + \sum_{r\in\mathcal{R}}\sigma_{ir}\hat{r}_{r}\qquad\forall{i}\in\mathcal{M}\]

where \(D_{s}\) is called the dilution rate for stream \(s\in\mathcal{S}\); the dilution rate has units of inverse time. Finally, at steady-state, all accumulation terms vanish, giving:

\[\sum_{s\in\mathcal{S}}v_{s}C_{s,i}D_{s} + \sum_{r\in\mathcal{R}}\sigma_{ir}\hat{r}_{r} = 0\qquad\forall{i}\in\mathcal{M}\]

If a reactor is well mixed, then the concentration of component \(i\in\mathcal{M}\) in the reactor is the same as the concentration of component \(i\in\mathcal{M}\) in the exit streams.

Definition 33 (Well mixed concentration balance)

Suppose we have a reaction set \(\mathcal{R}\) involving the species set \(\mathcal{M}\) that is occurring in a well-mixed chemical reactor with stream set \(\mathcal{S}\); partition the streams into the subsets \(\mathcal{S}^{+}\) and \(\mathcal{S}^{-}\), where \(\mathcal{S}^{+}\) denotes the subset of streams that enter the reactor, and \(\mathcal{S}^{-}\) represents the subset of streams that exit the reactor.

Then, the constant volume well-mixed concentration balance for component \(i\in\mathcal{M}\) is given by:

(76)#\[\frac{dC_{i}}{dt} = \sum_{s\in\mathcal{S^{+}}}v_{s}C_{s,i}D_{s} - C_{i}\sum_{s\in\mathcal{S}^{-}}v_{s}D_{s} + \sum_{r\in\mathcal{R}}\sigma_{ir}\hat{r}_{r}\qquad\forall{i}\in\mathcal{M}\]

where \(C_{i}\) denotes the concentration of component \(i\in\mathcal{M}\) in the reactor, and the kinetic model for the reaction rate \(\hat{r}_{r}\) is evaluated at composition \(C_{i}\).

At steady-state, the accumulation terms in Eqn. (76) are zero, which gives the well-mixed steady-state constant volume concentration balance:

(77)#\[\sum_{s\in\mathcal{S^{+}}}v_{s}C_{s,i}D_{s} - C_{i}\sum_{s\in\mathcal{S}^{-}}v_{s}D_{s} + \sum_{r\in\mathcal{R}}\sigma_{ir}\hat{r}_{r} = 0\qquad\forall{i}\in\mathcal{M}\]

Estimating the steady-state well-mixed exit concentration#

At steady-state, the well-mixed concentration balances go from Ordinary Differential Equations (ODEs) to a system of non-linear algebraic equations, depending upon the function form of the reaction kinetics. In the general case, estimating the steady-state exit composition is a challenging problem. We can approximate the solution to this problem as an optimization problem.

To estimate the steady-state well-mixed concentration, we need to solve an optimization problem, i.e., we need to search for exit concentrations that make our steady-state concentration balances zero. We can do this via the Optim.jl package. The problem we are solving is to find a concentration vector \(x\) that makes the residual vector \(\epsilon\) small:

\[\min_{x}\epsilon^{T}\epsilon\]

subject to the constraints on the concentration \(0\leq{x}\leq\infty\), where the residual for component \(i\in\mathcal{M}\) is defined as:

\[\epsilon_{i} \equiv \sum_{s\in\mathcal{S^{+}}}v_{s}C_{s,i}D_{s} - x_{i}\sum_{s\in\mathcal{S}^{-}}v_{s}D_{s} + \sum_{r\in\mathcal{R}}\sigma_{ir}\hat{r}_{r}(x)\qquad\forall{i}\in\mathcal{M}\]

Various techniques exist to search for good concentration vectors \(x\). We’ll use a derivative-free search method called Nelder-Mead to generate candidate values for the concentration vector \(x\); we’ll keep generating guesses and checking their residual values until we find a candidate solution that meets some smallness criteria.

Example#

Summary#

In this lecture, we: