

InterviewSolution
Saved Bookmarks
1. |
Show that in a first order reaction, time pequired for fompletion of `99.9%` is 10 times of half-life `(t_(1//2))` of the reaction. |
Answer» When reaction is completed `99.9%, [R]_(n)=[R]_(0)-0.999[R]_(0)` `k=(2.303)/(t)log ""([R]_(0))/([R])` `=(2.303)/(t )log ""([R]_(0))/([R]_(0)-0.999[R]_(0))=(2.303)/(t)log 10^(3)` `t=6.909//K` ltbr For half-life of the reaction `t_(1//2) =0.693//K` `(t)/(t_(1//2))=(6.909)/(k)xx(k)/(0.693)=10` |
|