Question

( t_{1 / 2}=60 mathrm{min} )
Half life period ( =t_{1 / 2}=frac{0.693}{k} )
[
begin{array}{c}
k=text { rate constant }
60=frac{0.693}{k}
k=0.01155 mathrm{min}
end{array}
]
( k=frac{2 cdot 303}{t} log left[frac{left.a_{0}right]}{[a]}- ) Inition right.
After consuming 90"/ final conc. Dill be ( 10^{circ}(cdot ) of initial conc.
Initial ( operatorname{con} c=x )
Find ( operatorname{con} c cdot i 8 cdot 10^{circ} cdot 0 . f x=frac{10}{100} x=0.1 x )
( frac{text { Initiol } cos c}{text { Find } operatorname{conc}}=frac{x}{0.1 x}=10 )
( 0.01155=frac{2.303}{t} log (10) )
( 199 cdot 4 ) minuts
[
t=1
]

# SICUIT AI IWC =V.uss aun Example 23. The half life of a first order reaction is 60 min. How long will it take to consume 90% of the reactant?

Solution