A batch reactor converts component A into B, which in turn decomposes into C :
\stackrel{k_{1}}{ A \rightarrow B } \stackrel{k_{2}}{\rightarrow C }
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{array}{ll}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{array}
Find the constant temperature that maximizes the amount of B, for 0 \leq t \leq 6 min.
