`Ararr` Product, `[A]_(0) = 2M`. After `10 min` reaction is `10%` completed. If `(d[A])/(dt) = k[A]`, then `t_(1//2)` is approximately

A first order reaction is 10% complete in 20 min. the time taken for 19% completion is :

Consider the reaction : A rarr B + C . Initial concentration of A is 1 M. 20 minutes time is required for completion of 20% reaction. If (d[B])/(dt) =k[A], then calculate half life (t_(1//2)) of reaction.

For the following reaction, A overset(500 K)rarr Products A overset(400K//"catalyst")rarr Product If ((-d[A])/(dt))_(500 K)=-((d[A])/(dt))_(400 K) no catalyst in presence of catalyst Then E_(a) is approximately :

For the following reaction, A overset(500 K)rarr Products A overset(400K//"catalyst")rarr Product If ((-d[A])/(dt))_(500 K)=-((d[A])/(dt))_(400 K) no catalyst in presence of catalyst Then E_(a) is approximately :

If 60% of a first order reaction was completed in 60 min, 50% of the same reaction would be completed in approximately

A first order reaction is 20% completed in 10min, (70+h) min it take to complete 80% .What is h? (approximately) (log1.25=0.0969 )

For a reation: A rarr Product, rate law is -(d[A])/(dt)=K[A]_(0) . The concentration of A left after time t when t=1/K is:

In the reaction 3Ararr2B , rate of reaction +(d[B])/(dt) is equal to :