(Challenging) Cubic autocatalytic steps are important in a reaction mechanism
referred to as the "brusselator"
(named in honor of the research group in Brussels that initially discovered
this mechanism): $$\begin{array}{l}
\mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{X} \\
2 \mathrm{X}+\mathrm{Y} \stackrel{k_{2}}{\longrightarrow} 3 \mathrm{X} \\
\mathrm{B}+\mathrm{X} \stackrel{k_{3}}{\longrightarrow} \mathrm{Y}+\mathrm{C}
\\\
\mathrm{X} \stackrel{k_{4}}{\longrightarrow} \mathrm{D}
\end{array}$$
If \([\mathrm{A}]\) and \([\mathrm{B}]\) are held constant, this mechanism
demonstrates interesting oscillatory behavior that we will explore in this
problem.
a. Identify the autocatalytic species in this mechanism.
b. Write down the differential rate expressions for \([\mathrm{X}]\) and
\([\mathrm{Y}]\)
c. Using these differential rate expressions, employ Euler's method (Section
35.6 ) to calculate \([\mathrm{X}]\) and \([\mathrm{Y}]\) versus time under the
conditions \(\mathrm{k}_{1}=1.2 \mathrm{s}^{-1}, \mathrm{k}_{2}=0.5
\mathrm{M}^{-2} \mathrm{s}^{-1}, \mathrm{k}_{3}=\)
\(3.0 \mathrm{M}^{-1} \mathrm{s}^{-1}, \mathrm{k}_{4}=1.2 \mathrm{s}^{-1},\) and
\([\mathrm{A}]_{0}=[\mathrm{B}]_{0}=1 \mathrm{M} .\) Begin
with \([\mathrm{X}]_{0}=0.5 \mathrm{M}\) and \([\mathrm{Y}]_{0}=0.1 \mathrm{M} .\)
A plot of \([\mathrm{Y}]\) versus
\([\mathrm{X}]\) should look like the top panel in the following figure.
d. Perform a second calculation identical to that in part (c), but with
\([\mathrm{X}]_{0}=3.0 \mathrm{M}\) and \([\mathrm{Y}]_{0}=3.0 \mathrm{M} .\) A
plot of \([\mathrm{Y}]\) versus
\([\mathrm{X}]\) should look like the bottom panel in the following figure.
e. Compare the left and bottom panels in the following figure. Notice that the
starting conditions for the reaction are different (indicated by the black
spot). What DO the figures indicate regarding the oscillatory state the system
evolves to?
