Answer:
Half-life of the reaction is 0.072s.
Explanation:
In a first-order reaction, half-life, t1/2, is defined as:
[tex]t_{1/2} = ln2 / k[/tex] (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):
[tex]t_{1/2} = ln2 / 9.6s^{-1}[/tex]
[tex]t_{1/2} = 0.072s[/tex]
I hope it helps!
