Question

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

Answer 1

Answer:

Cyclopropane will take $0.072 secs$ to reach $0.50$ of its initial concentration.

Explanation:

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

half-life = $\frac{0.693}{k}$

Half-life of cyclopropane will be,

$t_{1/2}= \frac{0.693}{9.6s^{-1} }$

$t_{1/2} =0.072 sec$

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