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?

Answer 1

Half-life of the reaction is 0.072s.

Step-by-step explanation:

In a first-order reaction, half-life, t1/2, is defined as:

`t_(1/2) = ln2 / k` (1)

Where k is the rate constant of the reaction.

In the problem, the thermally descomposition of cyclopropane has a rate constant of 9.6s⁻¹. Replacing in (1):

`t_(1/2) = ln2 / 9.6s^(-1)`

`t_(1/2) = 0.072s`

I hope it helps!