A batch reactor converts component A into B, which in turn decomposes into C :
\begin{gathered}k_{1} \quad k_{2} \\A \rightarrow B \rightarrow C\end{gathered}
where k_{1}=k_{10} e^{-E_{1} / R T} and k_{2}=k_{20} e^{-E_{2} / R T}.
The concentrations of A and B are denoted by x_{1} and x_{2}, respectively. The reactor model is
\begin{aligned}&\frac{d x_{1}}{d t}=-k_{10} x_{1} e^{-E_{1} / R T} \\&\frac{d x_{2}}{d t}=k_{10} x_{1} e^{-E_{1} / R T}-k_{20} x_{2} e^{-E_{2} / R T}\end{aligned}
Thus, the ultimate values of x_{1} and x_{2} depend on the reactor temperature as a function of time. For
\begin{aligned}k_{10} &=1.335 \times 10^{10} min ^{-1}, & & k_{20}=1.149 \times 10^{17} min ^{-1} \\E_{1} &=75,000 J / g mol , & & E_{2}=125,000 J / g mol \\R &=8.31 J /( g mol K ) & & x_{10}=0.7 mol / L , \quad x_{20}=0\end{aligned}
Find the constant temperature in K that maximizes the amount of B, for 0 \leq t \leq 8 min. Next allow the temperature to change as a cubic function of time
T(t)=a_{0}+a_{1} t+a_{2} t^{2}+a_{3} t^{3}
Find the values of a_{0}, a_{1}, a_{2}, a_{3} that maximize x_{2} by integrating the model and using a suitable optimization method.
