InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
`99%` at a first order reaction was completed in `32 min`. When will `99.9%` of the reaction complete.A. 48 minB. 50 minC. 35 minD. 39 min |
|
Answer» Correct Answer - A Time required to complete any particular fraction of first order reaction is `_(t//2)=(2.303)/(k)log(1)/(1-f)` therefore `t_(99%)=(2.303)/(k)log(1)/(1-0.99)` `=(2.303)/(k)log(1)/(0.01)` `=(2.303)/(k)log10^(2)` `t_(99.9%)=(2.303)/(k)log(1)/(1-0.999)` `=(2.303)/(k)log(1)/(0.001)` `=(2.303)/(k)log10^(3)` Dividing Eq. `(1)` by Eq. `(2)` , we get `(t_(99%))/(t_(99.9%))=(log10^(2))/(log10^(3))=(2log10)/(3log10)=(2)/(3)` Thus, `t_(99.9%)=(3)/(2)t_(99%)` `=(3)/(2)(32min)=48min` |
|