Discussion Forum

The first-order decomposition of $H_{2}O_{2}$ in an appropriate medium is characterised by a rate constant of 0.2303 min^-1 . What is the time  (in min) required to complete 9/10 fraction of the reaction?

Answer:

Explanation:

Given, rate constant (k) for 

first order=0.2303 min^-1

 For first  order reaction,

 Time ,$(t)= \frac{2.303}{k}\log \left[\frac{a}{a-x}\right]$  ...........(i)

 where, a= initial  concentration

(a-x)= concentration at time (t)

 x= amount complete

 let, initial concentration (a)=1

                               x=$\frac{9}{10}$

 then , (a-x)=$1-\frac{9}{10}=\frac{1}{10}$

 From equation (i),

$t= \frac{2.303}{0.2303}\log\frac{1}{\left(\frac{1}{10}\right)}$

  t= 10 log 10

 $t= 10 \times 1=10 min^{-1}$

 Hence, option (b) is correct