Answer:
a) 79.66 seconds is the half-life of the reaction.
b) It will take [tex]4.776\times 10^1 [/tex] seconds for 34% of a sample of an acetone to decompose.
c) It will take [tex]2.537\times 10^2 [/tex] seconds for 89% of a sample of an acetone to decompose.
Explanation:
The decomposition of acetone follows first order kinetics
The rate constant of the reaction = k = [tex]8.7\times 10^{-3} s^{-1}[/tex]
a)
Half life of the reaction = [tex]t_{1/2}[/tex]
For the first order kinetic half life is related to k by :
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
[tex]t_{1/2}=\frac{0.693}{8.7\times 10^{-3} s^{-1}}=79.66 s=7.966\times 10^1 s[/tex]
79.66 seconds is the half-life of the reaction.
b)
Let the initial concentration of acetone be = [tex][A_o][/tex]
Final concentration of acetone left after t time = [A]
[tex]A=(100\%-34\%)[A_o]=66\%[A_o]=0.66[A_o][/tex]
For the first order kinetic :
[tex][A]=[A_o]\times e^{-kt}[/tex]
[tex]0.66[A_o]=[A_o]\times e^{-8.7\times 10^{-3} s^{-1}\times t}[/tex]
Solving for t;
[tex]t=47.76 s[/tex]
It will take 47.76 seconds for 34% of a sample of an acetone to decompose.
c)
Let the initial concentration of acetone be = [tex][A_o][/tex]
Final concentration of acetone left after t time = [A]
[tex]A=(100\%-89\%)[A-o]=11\%[A_o]=0.11[A_o][/tex]
For the first order kinetic :
[tex][A]=[A_o]\times e^{-kt}[/tex]
[tex]0.11[A_o]=[A_o]\times e^{-8.7\times 10^{-3} s^{-1}\times t}[/tex]
Solving for t;
[tex]t=2.537\times 10^2 s[/tex]
It will take [tex]2.537\times 10^2 [/tex] seconds for 89% of a sample of an acetone to decompose.