Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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

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

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App