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The half-life of a compound is 76 years for 42% of a given sample of the compound decomposes in 60 minutes.
The rate law for first order kinetics is
k = [tex]\frac{2.303}{t}log\frac{a}{a-x}[/tex]
Where
k - rate constant
t - time passed by the sample
a - initial amount of the reactant
a-x - amount left after the decay process
The expression for the half-life of the compound is
[tex]t_{\frac{1}{2} }=\frac{0.693}{k}[/tex]
Where k is the rate constant
It is given that
t = 60 minutes
a = 100 g
a-x = 100 g - 42 g = 58 g
(Since it is given that 42% of the given compound decomposes)
Then,
k = [tex]\frac{2.303}{60}log\frac{100}{58}[/tex]
= 9.08 × [tex]10^{-3}[/tex] [tex]years^{-1}[/tex]
Then, the half-life of the compound is
[tex]t_{\frac{1}{2} }=\frac{0.693}{k}[/tex]
= [tex]\frac{0.693}{9.08*10^{-3} }[/tex]
= 76.3 years ≅ 76 years
Therefore, the half-life of the given compound is 76 years.
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