Home
Class 12
CHEMISTRY
In a reaction A rarr products , when st...

In a 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 `:`

A

zero

B

one

C

two

D

0.5

Text Solution

Verified by Experts

The correct Answer is:
C

`((t_(1//2))_(1))/((t_(1//2))_(2))=((a_(2))/(a_(1)))^(n-1) rArr (120)/(240) =((4xx10^(-2))/(8xx10^(-2)))^(n-1) rArr n=1=1 rArr n=2`
Promotional Banner

Similar Questions

Explore conceptually related problems

The half period for the decomposition of a compound is 20 minutes. If the initial concentration is made twice, the half life period becomes 10 minutes. Calculate the order of reaction.

The half life period of decomposition of a compound is 50 minutes. If the initial concentration is halved, the half life period is reduced to 25 minutes. What is the order of reaction ?

A first order reaction is found to have a rate constant, k = 4.2 xx 10^(-12) s^(-1) . Find the half-life of the reaction.

The half-life period of a first order reaction is 10 minutes. Starting with initial concentration 12 M, the rate after 20 minutes is

The rate of the reaction A rarr products, at the initial concentration of 3.24 xx 10^(-2)M is nine times its rate at another initial concentration of 1.2 xx 10^(-3)M . The order of reaction is

The half-life of a reaction is halved as the initial concentration of the reaction is doubled. The order of the reaction is