Chemical reaction kinetics: Introduction
Simulation of first-order kinetics
In case of first-order kinetics of the reaction \[\text{A}\longrightarrow \text{B}\] the differential equation that describes the time course of the concentration of substance A can be solved for various parameter values (initial concentrations and reaction rate constant \(k\)) and graphs of solutions can be computed and plotted.
In the following simulation, the time course of the chemical reaction is simulated, and the graphs of the concentrations \([\text{A}]\) and \([\text{B}]\) of substance A and B, respectively, are plotted. You can vary the initial concentrations and rate constants, and see what the effect of this variation. So you can find out for example that a smaller rate constant only means that the conversion goes slower.
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Time course of the concentration of the product We can also determine the time course of the concentration of the produced substance B in the chemical reaction \(\text{A}\longrightarrow \text{B}\), with reaction rate constant \(k\). Let the concentration of substance A and B at time \(t\) be denoted as \(x(t)\) and \(y(t)\), respectively. In first-order kinetics we have: \[\frac{\dd x}{\dd t}=-k\, x\] This is a differential equation of an exponential decay with the solution \[x(t)=x_0 e^{-kt}\quad\text{with}\quad x_0=x(0)\] Let \(y(0)=y_0\) and \(x_0+y_0=a\), for some constant \(a\), then we have at any time \[x(t)+y(t)=a\] and thus also \[x'(t)+y'(t)=0\] It follows that \[y'(t)=-x'(t)=k\, x(t)=k\bigl(a-y(t)\bigr)\] and \[y(t)=a - x_0 e^{-kt}\text.\] The differential equation \[y'(t)=k\bigl(a-y(t)\bigr)\] describes limited exponential growth and its general solution is \[y(t)=a - c\, e^{-kt}\] for some constant \(c\).
