Answer:
[tex]t_{1/2}= 20.47\ minute[/tex]
Explanation:
given,
constant rate for the reaction(k) = 5.64 × 10⁻⁴ s⁻¹
to calculate the half life = ?
[tex]t_{1/2} = \dfrac{0.693}{k}[/tex]
[tex]t_{1/2} = \dfrac{0.693}{5.64\times 10^{-4}}[/tex]
on solving the above equation
[tex]t_{1/2} = 1228.72 s[/tex]
[tex]t_{1/2} = \dfrac{1228.72}{60}[/tex]
[tex]t_{1/2}= 20.47\ minute[/tex]
hence, the half-life of cyclopropane at this temperature =[tex]t_{1/2}= 20.47\ minute[/tex]
