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For the given sequential reaction $A\;\underrightarrow{k_1}\;B\;\underrightarrow {k_2}C$ the concentration of A,B & C at any time 't' is given by $[A]_t=[A]_oe^{-k_1t};[B]_t=\large\frac{k_1[A_o]}{k_2-k_1}$$[e^{-k_1t}-e^{-k_2t}];[C]_t=[A_o]-([A]_t-([A]_t+[B]_t)$.The time at which a concentration of B is maximum is

$\begin{array}{1 1}(a)\;\large\frac{k_1}{k_2-k_1}&(b)\;\large\frac{1}{k_2-k_1}\normalsize ln \large\frac{k_1}{k_2}\\(c)\;\large\frac{1}{k_1-k_2}\normalsize ln \large\frac{k_1}{k_2}&(d)\;\large\frac{k_2}{k_2-k_1}\end{array}$

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Answer:$\;\large\frac{1}{k_1-k_2}\normalsize ln \large\frac{k_1}{k_2}$
By putting $\large\frac{d[B]}{dt}=$$0$, We get $t_{max}=\large\frac{1}{k_1-k_2}$$ln \large\frac{k_1}{k_2}$
answered Dec 16, 2013 by sreemathi.v
edited Jul 26, 2014 by balaji.thirumalai

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