Chemical Kinetics and Nuclear ChemistryHard

Question

Consider the following consecutive first-order reaction. $A\overset{\quad K_{1}\quad}{\rightarrow}B\overset{\quad K_{2}\quad}{\rightarrow}C$

If K1 = 0.01 min−1 and K1: K2 = 1:2, after what time from the start of reaction, the concentration of ‘B’ will be maximum? (ln 2 = 0.7)

Options

A.70 min
B.140 min
C.35 min
D.700 min

Solution

${t\frac{\ln\left( \frac{K_{2}}{K_{1}} \right)}{K_{2} - K_{1}}\frac{\ln\left( \frac{2}{1} \right)}{0.02 - 0.01}\text{ min}}_{\max}$

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