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

The half life period of a substance is 50 minutes at a certain initial

1.	The half life period of a substance is 50 minutes at a certain initial concentration. When the concentration is reduced to one half of its initial concentration, the half life period is found to be 25 minutes. Calculate the order of reaction.
Answer» Solution :Suppose theintial 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 orderof 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*"`.