Prove that `t_(75) = 2 t_(50)` for `1^(st)` order reaction.

Prove that t_(99.9%)// t_(50)% = 10 for a first order reaction?

The value of t_(0.875)/(t_(0.50) for n^(th) order reaction is

Derive the integrated rate law equation for 1 ^(st) order reaction and write its one use.

The unit of 1^(st) order rate constant are

For a partiocular reaction with initial conc. Of the rectents as a_(1) and a_(2) , the half-life period are t_(1) and t_(2) respectively. The order of the reaction (n) is given by :

The plot of [R] (concentration) versus t (time) for a zero order reaction is

If t_(1//2) is 0.693 sec. for a first order reaction. Calculate reaction rate constant.

The unit of 1 s t order rate constant are

The integrated rate equations can be fitted with kinetic data to determine the order of a reaction. The integrated rate equations for zero, first and second order reactions are : Zero order : [Al =-kt+ [A]_0 First order : log [A] = -(kt)/2.303 + log [A]_0 Second order : 1/([A])=kt+1/[A]_0 These equations can also be used to calculate the halt life periods of different reactions, which give the time during which the concentration of a reactant is reduced to half of its initial concentration, i.e, at time t_(1//2), [A] = [A]_0//2 Answer the following (1 to 5) question : For a second order reaction, the correct plot of t_(1//2) vs. 1//[A]_0 is