Home
Class 12
CHEMISTRY
In the reaction A rarr Products, when st...

In the reaction `A rarr` Products, when start is made from `8.0 xx 10^(-2)M` of A, half-life is found to be 120 minute. For the initial concentration `4.0 xx 10^(-2)M`, the half-life of the reaction becomes 240 minute. The order of the reaction is :

Text Solution

Verified by Experts

The correct Answer is:
2

`((t_(1//2))_(1))/((t_(1//2))_(2)) = ((a_(2))/(a_(1)))^(n-1), (120)/(240) = ((4 xx 10^(-2))/(8 xx 10^(-2)))^(n-1), n = 2`
Promotional Banner

Similar Questions

Explore conceptually related problems

If 'a' is the initial concentration of the reactant, the half-life period of the reaction of nth order is inversely proportional to:

A first order reaction is found to have a rate constant, k=5.5xx10^(-4)s^(-1) .Find out the half-life of the reaction.

Half change time for a 1st order reaction is 20 minutes, hence the rate constant for the reaction is

Rate constant (k)of a reaction is 5xx10^(-2)s^(-1) .Find the half life (t_(1/2)) of the reaction.

If half-life of a reaction Is directly proportional to Initial concentration of the reactant, what Is the order of the reaction?

A first order reaction has spećfic reaction rate 2 (min)^(-1) . The half life of the reaction' will be

Consider a certain reaction ArarrProducts with k=2.0 xx 10^(-2) (s)^(-1) .Calculate the concentration of A remaining after 100s if the initial concentration of A is 1.0 molL^(-1)