Question

Derive the relation between half life and rate constant for a first order reaction

Answer 1

Try Yourself!! Its very easy for you but hard for me !!!!!!!!!!!!

Answer 2

Answer: $A\rightarrow Products$

$rate=k[A]$    (1)

also $rate=-\frac{d[A]}{dt}$    (2)

equating 1 and 2  

$k[A]=-\frac{d[A]}{dt}$    

$k\times dt=-\frac{d[A]}{[A]}$  

Integrating both sides and putting the limits for time as t=0 to t=t and for concentration from A=$[a_0]$ to A=a  

$k(t-0)={ln[A_0]-ln[A]$

Rate law expression for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{[a_0]}{[a]}$

where,

k = rate constant

t = time taken for decay process

$a_0$ = initial amount of the reactant

a = amount left after decay process

Half life is the time taken by the reactants to decompose to half of its value.

Thus $t=t_\frac{1}{2}$ and $a=\frac{a_0}{2}$

$t_\frac{1}{2}=\frac{2.303}{k}\log\frac{[a_0]}{[a_0/2]}$

$t_\frac{1}{2}=\frac{0.693}{k}$