Answer:
[tex]t_{1/2}=30.14\ min[/tex]
Explanation:
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given:
75 % is decomposed which means that 0.75 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.75 = 0.25
t = 60 min
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.25=e^{-k\times 60}[/tex]
k = 0.023 min⁻¹
Considering the expression for half life as:-
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]t_{1/2}=\frac{\ln 2}{0.023}\ min[/tex]
[tex]t_{1/2}=30.14\ min[/tex]
The half-life of a compound if 75% of a given sample of the compound decomposes in 60 min - 30 min
Given:
t = time of decomposition = 60 minutes
a = let initial amount of the reactant = 100g
a - x = amount left after decay process = (100-75)g = 25g
Solution:
Expression for rate law for first-order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
Placing the given values:
[tex]60=\frac{2.303}{k}\log\frac{100}{25}\\\\k=0.023min^{-1}[/tex]
Half-life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}t_{\frac{1}{2}}=\frac{0.693}{0.023}\\=30\ min[/tex]
Thus, the half-life of a compound if 75% of a given sample of the compound decomposes in 60 min - 30 min
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