logo

Ask Questions, Get Answers

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

The decomposition of hydrogen bromide into hydrogen and iodine is second order reaction.The reaction rate constant is $K=0.08Lmol^{-1}s^{-1}$.How long it take an initial concentration of 0.04M to decrease to half this concentration?

$\begin{array}{1 1}(a)\;300sec&(b)\;315sec\\(c)\;212.5sec&(d)\;312.5sec\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
$2HBr\rightarrow H_2+Br_2$
At $t=0$ $2HBr\Rightarrow a$
At $t=t$ $2HBr\Rightarrow a-x,H_2\Rightarrow x/2,Br_2\Rightarrow x/2$
Rate $=\large\frac{dx}{2dt}$$=K(a-x)^2$
$\large\frac{dx}{(a-x)^2}$$=2Kdt$
$\large\frac{dx}{(a-x)^2}$$=K'dt$
$2K=K'$
$K'$=rate constant=0.08$Lmol^{-1}s^{-1}$
$-\large\frac{1}{a-x}\bigg]_0^{a/2}$$=K't\big]_0^{t_{1/2}}$
$-\big[\large\frac{1}{a}-\frac{1}{a/2}\big]$$=k't_{1/2}$
$t_{1/2}=\large\frac{1}{k'a}$
$t_{1/2}=\large\frac{1}{0.08\times 0.04}$
$t_{1/2}=312.5sec$
Hence (d) is the correct answer.
answered Dec 18, 2013 by sreemathi.v
 

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
...