Reaction kinetics of the chemical reaction A → B

Unlimited growth: Applications of exponential growth models

Reaction kinetics of the chemical reaction <br/> A → B

A biological process is often very complex and to understand it we usually introduce a simplified model. Since we are interested in concentrations of proteins, ions, and other substances, as well as in the changes of these concentrations, is the model of the change process, also known as a kinetic model, usually a differential equation or a system of differential equations.

We illustrate first kinetic modeling using a simple chemical reaction in which A molecule is converted into a molecule B: \[\text{A}\longrightarrow \text{B}\]

Reaction rate constant We assume such a small time interval \({\Delta}t\) that a molecule A is either converted during this interval into molecule P or remains intact. Suppose that the probablity of conversion during this interval is equal to \(k\cdot {\Delta}t\), for a certain reaction rate constant \(k\). The constant \(k\) therefore has a unit \(\textit{time}^{-1}\) and by definition we have: \({\Delta}t\le \frac{1}{k}\). In reality, this description of the chance of conversion is better when \({\Delta}t\) gets smaller. In practice we will therefore demand: \({\Delta}t\ll \frac{1}{k}\).

This introduction of the reaction rate constant \(k\) is actually based on a description of the chemical reaction as a so-called Markov chain. The foundational characteristic properties are:

The conversions of molecules A happen randomly in time;
The probability of a conversion of a molecule A in the time interval \([t, t+{\Delta}t]\) does not depend on what has happened prior to the interval of interest;
When circumstances do not change, then the probability of a conversion of A during the time interval \([t, t+{\Delta}t]\) does not depend on the time \(t\), but only on the length of the time interval \({\Delta}t\).

The differential equationfor first-order reaction kinetics We now continue with the kinetic modelling of the reaction \(\text{A}\longrightarrow \text{B}\).

Suppose that \(n_\text{A}(t)\) is the amount of the reagent A at time \(t\), that is, the number of molecules. Then the expected decrease in the number of molecules A in the time interval \([t, t+{\Delta}t]\) is equal to\(n_\text{A}(t)\cdot k\cdot{\Delta}t\). This leads to \[n_\text{A}(t+{\Delta}t)-n_\text{A}(t)=-n_\text{A}(t)\cdot k\cdot{\Delta}t\] In other words: \[\frac{n_\text{A}(t+{\Delta}t)-n_\text{A}(t)}{{\Delta}t}=- k\cdot n_\text{A}(t)\] When we choose \({\Delta}t\) smaller and smaller, then we will get in the limiting case \({\Delta}t\rightarrow 0\) the equation \[\frac{\dd n_\text{A}}{\dd t}=- k\cdot n_\text{A}\] This is the differential equation of first-order reaction kinetics, because the instantaneous change of the number of molecules of A is proportional to this number at the given time. In mathematical language: \(n_\text{A}'(t)=-k\cdot n_\text{A}(t)\).

Usually it is more convenient to use concentrations instead of volumes. Suppose the volume \(V\) is not changed during the chemical reaction. Then the concentration \(C_\text{A}\) of substance A in a volume \(V\) is given by \(C_\text{A}=n_\text{A}/V\) and we get the following differential equation: \[\frac{\dd C_\text{A}}{\dd t}=-k\cdot C_\text{A}\]

Chemists write the concentration of a substance A with [A]. Then: \[\frac{d[\text{A}]}{\dd t}=-k\cdot [\text{A}]\]

Concentration-time profile in first-order kinetics Let \(x(t)\) be the concentration of substance A at time \(t\) in the chemical reaction \(\text{A}\longrightarrow \text{B}\). In case of first-order reaction kinetics we have \[\frac{\dd x}{\dd t}=-k\, x\] This is a differential equation of a exponential decay with the following solution: \[x(t)=x_0 e^{-kt}\quad\text{with}\quad x_0=x(0)\]

For the math enthusiast we present (again) the proof via the method of an integrating factor. The so-called method of separation of variables is actually more commonly used.

The differential equation\[\frac{\dd x}{\dd t}=-k\, x\] can be rewritten ad \[\frac{\dd x}{\dd t}+k\, x=0\] Multiplying both sides by \(e^{kt}\) leads to \[e^{kt}\frac{\dd x}{dt}+ke^{kt}x=0\] and according to the product rules for differentiating this is equivalent to \[\frac{\dd}{\dd t}\left(e^{kt}x(t)\right)=0\] When the derivative of a function is equal to zero, then the function itself is constant, say equal to \(c\). So \[e^{kt}x(t)=c\] In other words: \[x(t)=c\, e^{-kt}\] At \(t=0\) we have \(x(0)=c\, e^0=c\) So: \[x(t)=x_0 e^{-kt}\quad\text{with}\quad x_0=x(0)\]