$\begin{array}{1 1}(a)\;14.23\%&(b)\;0\%\\(c)\;6.25\%&(d)\;50\%\end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

As the given reaction is first order reaction.

$K=\large\frac{1}{t}$$ln\large\frac{[A_0]}{[A_t]}$

$K=\large\frac{1}{1hr}$$ln\large\frac{[A_0]}{25\%[A_0]}$

$\;\;\;\;=ln 4hr^{-1}$

Now,as K remains constant for a particular reaction,

$K=\large\frac{1}{2hr}$$ln\large\frac{[A_0]}{[A_t]}$$=ln 4hr^{-1}$

$ln\large\frac{[A_0]}{[A_t]}$$=2ln \;4hr^{-1}$

$ln\large\frac{[A_0]}{[A_t]}$$=ln \;16hr^{-1}$

$\large\frac{[A_0]}{[A_t]}$$= \;16$

$[A_t]=\large\frac{[A_0]}{16}$

$[A_t]=(6.25\%)[A_0]$

$\large\frac{[A_t]}{[A_0]}$$= 6.25\%$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...