Home
Class 12
CHEMISTRY
How much time is requred for two - third...

How much time is requred for two - third completion of a first order reaction having, `K=5.48xx10^(-14)S^(-1)`?

A

`2.01xx10^(11) s`

B

`~2.01xx10^(13)s`

C

`8.08xx10^(13) s`

D

`16.04xx10^(11)` s

Text Solution

Verified by Experts

The correct Answer is:
B
`kt_(2//2)=ln ((C_(o))/(C_(o) -(2)/(3) C_(o)))=ln3, t_(2//3) =(ln3)/(5.48xx10^(-4))~~ 2.01xx10^(13)` sec
Promotional Banner

Similar Questions

Explore conceptually related problems

Find out two-third (2//3) life of a first order reaction in which k=5.48 xx 10^(-14)s^(-1)

The half life of a first order reaction having rate constant k=1.7xx10^(-5) sec^(-1) is :

Find the two-thirds life (t_(2//3)) of a first order reaction in which k=5.48xx10^(-1) "sec"^(-1) ( log 3 = 0.4771 , log 2 = 0.3010)

(a) For a reaction ABrarrP , the rate law is given by, r=k[A]^(1//2)[B]^(2). What is the order of this reaction? (b) A first order reaction is found to have a rate constant k=5.5xx10^(-14)s^(-1) , Find the half life of the reaction.

Show that time required to complete 99.9% completion of a first order reaction is 1.5 times to 99% completion.