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

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

Create a free account to view solution

View Solution Free
Topic: Chemical Kinetics and Nuclear Chemistry·Practice all Chemical Kinetics and Nuclear Chemistry questions

More Chemical Kinetics and Nuclear Chemistry Questions

The rate of a chemical reaction doubles for every 10oC rise of temperature. If the temperature is raised by 50oC, the ra...In most case, for a rise of 10K temperature the rate constant is doubled to tripled. This is due to the reason that...Saponification of ethyl acetate (CH3COOC2H5) by NaOH (Saponification of ethyl acetate by NaOH is second order Rxn) is st...Which of the following is pseudo first-order reaction?...For a reaction 2A + B + 3C → D + 3E, the following data is obtained.Exp. No. Concentration in mole per litre Initial rat...