Question

In the first order reaction the initial concentration of the reactant was m/10 after 8 min 20 sec the concentration becomes m/100 what is the rate constant

Answer 1

Rate constant for this reaction is 0.0046 $sec^{-1}$.

Explanation:

The integrate rate constant for first order reaction is expressed by the following equation

                                         $k=\frac{2.303}{t}log_{10}\frac{a}{(a-x)}$  .......(i)

where  $k$ = rate constant for first order ,  $t$ = time

           $a$ = initial concentration at time t = 0 , $(a-x)$ = concentration at time t

In question given that

  $a=\frac{m}{10}$ ,   $(a-x)=\frac{m}{100}$

  t = 8 min 20 sec = 500 sec

Put the above value in equation (i)

                    $k=\frac{2.303}{500sec}log_{10}\frac{\frac{m}{10} }{\frac{m}{100} }$

                    $k=\frac{2.303}{500sec}log_{10} 10$                $(log10=1)$

                    $k=0.0046 sec^{-1}$