The half-life periof of a substance is `50 min` at a certain initial concentration. When the concentration is reduced to one-half of its initial concentration, the half-life periof is found to be `25 min`. Calculate the order of reaction.

Explanation

Suppose the intial concentration in the first case is `a" mol L"^(-1)`. Then
`[A_(0)]_(1)=a,(t_(1//2))_(1)=50" minutes"`
`[A_(0)]_(2)=(a)/(2),(t_(1//2))_(2)=25" minutes"`
We know that for a reaction of nth order, `t_(1//2)prop(1)/([A_(0)]^(n-1)):.((t_(1//2))_(1))/((t_(1//2))_(2))=([A_(0)]_(2)^(n-1))/([A_(0)]_(1)^(n-1))={([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)`
Substituting the values, we get `(50)/(25)=((a//2)/(a))^(n-1)" or "(2)/(1)=((1)/(2))^(n-1)=((2)/(1))^(1-n)`
or `" "1-n=1" or "n=0.` Hence, the reaction is of zero-order.
(b) Graphical method. Graphically, half-life method can be used to test the order of reaction as follows :
`t_(1//2)prop(1)/([A_(0)]^(n-1))" or "t_(1//2)=K(1)/([A_(0)]^(n-1))`
where K is a constant of proportionality (not the rate constant.)
Thus, for zero order, `t_(1//2)=K[A_(0)].` Hence, plot of `t_(1 //2)"vs"[A_(0)]" will be linear passing through the origin and having slope = K*"`.