Respuesta :
Answer: The half life of cyclopropane at 760 K is 28.99 minutes.
Explanation:
Rate law expression for first order kinetics is given by the equation:
[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]
where,
k = rate constant = ?
t = time taken for decay process = 6.8 minutes
[tex][A_o][/tex] = initial amount of the sample = 100 grams
[A] = amount left after decay process = (100 - 15) = 85 grams
Putting values in above equation, we get:
[tex]k=\frac{2.303}{6.8min}\log\frac{100}{85}\\\\k=2.39\times 10^{-2}min^{-1}[/tex]
The equation used to calculate rate constant from given half life for first order kinetics:
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
where,
[tex]t_{1/2}[/tex] = half life of the reaction = ?
k = rate constant of the reaction = [tex]2.39\times 10^{-2}min^{-1}[/tex]
Putting values in above equation, we get:
[tex]t_{1/2}=\frac{0.693}{2.39\times 10^{-2}min^{-1}}=28.99min[/tex]
Hence, the half life of cyclopropane at 760 K is 28.99 minutes.
