Be sure to answer all parts. Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to airplane glue and nail polish. At high temperatures, it decomposes in a first-order process to methane and ketene (CH2═C═O). At 600°C, the rate constant is 8.7 × 10−3 s−1. (a) What is the half-life of the reaction? Give your answer in scientific notation. 7.97 × 10 s (b) How long does it take for 34% of a sample of acetone to decompose? s (c) How long does it take for 89% of a sample of acetone to decompose? Give your answer in scientific notation. × 10 s

Respuesta :

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.

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