Kinetics of the simple chemical reaction A → B

Chemical reaction kinetics: Introduction

Kinetics of the simple chemical reaction A → B

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

Informal definition of a differential equation A differential equation is an equation in which besides one or more unknown functions also their derivatives are present, and of which the solution is a function too.

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First we illustrate kinetic modelling through a simple chemical reaction in which compound A is converted into a substance B (one molecule will actually be converted into the other): \[\text{A}\longrightarrow \text{B}\]

Reaction rate constant We assume such a small time interval \({\Delta}t\) that a molecule A in this interval is either converted into molecule B or remains intact. Suppose that the probability of conversion during this interval is equal to \(k\cdot {\Delta}t\), for a certain reaction rate constant \(k\). The rate 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 probability of conversion works better the more \({\Delta}t\) is 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 Markov chain. The characteristic properties they are based:

The conversions of molecules A take place 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 took place prior to this interval;
If circumstances do not change, then the probability of a conversion from A in the time interval \([t, t+{\Delta}t]\) does not depend on the time \(t\), but only on the length of the time interval \({\Delta}t\).

Differential equation corresponding to first-order 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 reactant \(A\) at time \(t\), that is, the number of molecules. Then the expected decrease in the number of molecules \(A\) in the time interval \([t, t+{\Delta}t]\) equals \(n_A(t)\cdot k\cdot{\Delta}t\). This gives \[n_\text{A}(t+{\Delta}t)-n_\text{A}(t)=-n_\text{A}(t)\cdot k\cdot{\Delta}t\] In other words \[\frac{n_\text{A}(t+{\Delta}t)-n_\text{A}(t)}{{\Delta}t}=- k\cdot n_\text{A}(t)\] If we now choose \({\Delta}t\) smaller and smaller, then we 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 corresponding to first-order kinetics, because the instantaneous change of the number of molecules of A is proportional to this number at any time. In mathematical language: \(n_\text{A}'(t)=-k\cdot n_\text{A}(t)\).

Usually it is convenient to use concentrations instead of amounts. Suppose that the volume \(V\) does not change during the chemical reaction. 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 denote the concentration of a substance A as [A]. Then: \[\frac{\dd[\text{A}]}{\dd t}=-k\cdot [\text{A}]\]

Time course of concentration in 1st order kinetics Let us denote the concentration of substance A at time \(t\) as \(x(t)\) for the chemical reaction \(\text{A}\longrightarrow \text{B}\). In case of first-order kinetics we have \[\frac{\dd x}{\dd t}=-k\, x\] This is a differential equation of exponential decay with the solution \[x(t)=x_0 e^{-kt}\quad\text{with}\quad x_0=x(0)\]

For the math enthusiast we give a proof via the method of an integrating factor. The method of separation of variables is actually more obvious, but requires knowledge and skills in integration of real functions.

We have \[\frac{\dd x}{\dd t}=-k\, x\tiny,\] that is, \[\frac{\dd x}{\dd t}+k\, x=0\] Multiplication by \(e^{kt}\) gives \[e^{kt}\frac{\dd x}{\dd t}+ke^{kt}x=0\] and according to the product rule for differentiation 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 \[e^{kt}x(t)=C\tiny,\] that is, \[x(t)=C\, e^{-kt}\tiny.\] If \(t=0\) then \(x(0)=C\, e^0=C\). So: \[x(t)=x_0 e^{-kt}\quad\text{with}\quad x_0=x(0)\]