InterviewSolution
Saved Bookmarks
| 1. |
What percentage of the initial concentration will react in 5 hours in a first order reaction whose rate constants is `5.78 xx 10^(-5)s^(-1)?`. |
|
Answer» For the first order reaction: `k=(2.303)/t log a/(a-x)` `k=5.78 xx 10^(-5)s^(-1), t=5hr=5 xx 60 xx 60=18000s` `log a/(a-x) = (kxxt)/(2.303)=(5.78 xx 10^(-5)s^(-1) xx 18000s)/(2.303)=0.4518` `(a/(a-x)) = "Antilog" 0.4518=2.83` Let `a=100%`, Therefore, `(100)/(100-x) =2.83` or `100 =283-2.83x` Let `a=100%`, Therefore, `(100)/(100-x)=2.83` or `100 = 283-2.83x` or `x=183/(2.83)=64.660 =64.66%` |
|