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Solution:- (a.) Considrring ist ordir readion:\begin{array}{l}X(-1)=1-\exp (-k \tau) \quad \text { idial PfR } \\0.99=1-\exp (-k c) \\-k \tau=\ln 0.01 \\k \tau=4.6\end{array}We are given with c k time.Being a pulse expriment we calculate J^{\prime}(t) firstJ^{\prime}(t) d t=\frac{c(t) d t}{\int_{0}^{\infty} c(t) d t}tione) \int_{0}^{\infty} c(t) d t=1416 \frac{(m g}{L} \times men ) \quad\{ calculated in excel py using aria undir the cwrery.Now \int_{0}^{t} J^{\prime}(t) d t=J(t) (faction of speies spending time b/w 0-t -(1) \min )Now from disporsion modelg(t)=\frac{1}{2}\left(1-\operatorname{erf}\left(\frac{1-t}{\sqrt{t}} \sqrt{P_{e}}\right)\right)Compasing 1 and 2 in excll by flotting we approkimate fit as,P e=0.05Now\begin{array}{c}\frac{C_{A}}{C_{A_{0}}}=\frac{4 a \exp \left(P_{\text {I }_{2}}\right)}{(1+a)^{2} \exp \left(\frac{\left.P_{C}\right)}{2}-(1-a)^{2} \exp \left(-\frac{a P_{C}}{2}\right)\right.} \\a=\sqrt{1+\frac{4 k C}{P_{C}}} \\k C=4.6 \\a=19.2 \\\frac{C_{A}}{C_{A 0}}=\frac{78.74}{659-205}=\frac{78.74}{454}=0.173\end{array}\begin{array}{l}X_{A}=\frac{C_{0}-C_{A}}{C_{A O}}=1-0.173=0.83 . \\X_{A}=0.83\end{array}dispirsion model.(b) Normal RTD\bar{x}=\int_{0}^{\infty} x(t) J^{\prime}(t) d tWhise x(t)=1-\exp (-k t) (batch reaita)\begin{aligned}k \tau & =4.6 \\\tau & =15.35 \\k & =\frac{4.6}{15.35} .\end{aligned}\tau=t_{\text {ang }}=\int_{0}^{\infty} J^{\alpha}(t) t d t (all calulated by aren undur curve in excel).1 \bar{x}=0.91through RTCaluilated in excel.(ce) Since we heure \tau=15.35 or k c=4.6 for ideal CSTK \& ist order\begin{aligned}x & =\frac{k c}{1+k \tau} \\& =\frac{4.6}{5.6} \\\bar{x} & =\frac{0.82}{C S T R}\end{aligned}We can also see that \bar{x} crR is similar to \bar{x} disfesion model with a very small peclet no as we knaw the lehaviour of model grien is similar to CSTR. ...