logo

Ask Questions, Get Answers

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

For the given sequential reaction $A\;\overset{k_1} \rightarrow \;B\;\overset{k_2} \rightarrow C$ 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 maximum concentration of $[B]$ is

$\begin{array}{1 1}(a)\;\large\frac{k_1[A_o]}{k_2-k_1}\bigg[\normalsize e^{\big(k_1/k_1-k_2\;ln\large\frac{k_1}{k_2}\big)}-e^{\big(k_2/k_1-k_2\;ln\large\frac{k_1}{k_2}\big)}\bigg]\\(b)\;\large\frac{k_1[A_o]}{k_2-k_1}\bigg[\normalsize e^{\big(k_1/k_2-k_1\;ln\large\frac{k_1}{k_2}\big)}-e^{\big(k_2/k_2-k_1\;ln\large\frac{k_1}{k_2}\big)}\bigg]\\(c)\;\large\frac{k_1[A_o]}{k_1-k_2}\bigg[\normalsize e^{\big(k_1/k_1-k_2\;ln\large\frac{k_1}{k_2}\big)}+e^{\big(k_2/k_1-k_2\;ln\large\frac{k_1}{k_2}\big)}\bigg]\\(d)\;\large\frac{k_1[A_o]}{k_1-k_2}\bigg[\normalsize e^{\big(k_1/k_1-k_2\;ln\large\frac{k_1}{k_2}\big)}-e^{\big(k_2/k_1-k_2\;ln\large\frac{k_1}{k_2}\big)}\bigg]\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Answer: $\large\frac{k_1[A_o]}{k_2-k_1}\bigg[\normalsize e^{\big(k_1/k_2-k_1\;ln\large\frac{k_1}{k_2}\big)}-e^{\big(k_2/k_2-k_1\;ln\large\frac{k_1}{k_2}\big)}\bigg]$
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}$
Where $[B]=\large\frac{k_1[A_o]}{(k_2-k_1)}$$[e^{-k_1t}-e^{-k_2t}]$
Putting $t_{max}$ for $t$ in [B] we get
$\large\frac{k_1[A_o]}{k_2-k_1}\bigg[\normalsize e^{\big(k_1/k_2-k_1\;ln\large\frac{k_1}{k_2}\big)}-e^{\big(k_2/k_2-k_1\;ln\large\frac{k_1}{k_2}\big)}\bigg]$
answered Dec 16, 2013 by sreemathi.v
edited Jul 26, 2014 by balaji.thirumalai
 

Related questions

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