Cyclopropane thermally decomposes by a first‑order reaction to form propene. If the rate constant is 9.6 s − 1 , what is the half‑life of the reaction? half‑life: s How long does it take for the concentration to reach 0.50 of its initial value? amount of time: s

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25-03-2020
Chemistry

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Cyclopropane thermally decomposes by a first‑order reaction to form propene. If the rate constant is 9.6 s − 1 , what is the half‑life of the reaction? half‑life: s How long does it take for the concentration to reach 0.50 of its initial value? amount of time: s

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mithun890 mithun890 · Answer 1 · 2020-03-30T00:02:15+02:00

Answer:

Cyclopropane will take [tex]0.072 secs[/tex] to reach [tex]0.50[/tex] of its initial concentration.

Explanation:

For a first order kinetics half- life of the reaction can be calculated by the formula,

half-life = [tex]\frac{0.693}{k}[/tex]

Half-life of cyclopropane will be,

[tex]t_{1/2}= \frac{0.693}{9.6s^{-1} }[/tex]

[tex]t_{1/2} =0.072 sec[/tex]

Since, [tex]0.50[/tex] means half of the initial concentration, which itself depicts the half- life of the cyclopropane. Therefore it, will take [tex]0.072 secs[/tex] to reach [tex]0.50[/tex] of its initial concentration i.e it will reduce to half of initial concentration in [tex]0.072 secs[/tex] when it thermally decomposes.