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)?`.

Explanation

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%`