Question

The isomerization of cyclopropane (CP) to propene (P) is a first order process. At 760 K, 15% of a sample of cyclopropane changes to propene in 6.8 minutes. What is the half- life of cyclopropane at 760 K

Answer 1

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:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant  = ?

t = time taken for decay process = 6.8 minutes

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process =  (100 - 15) = 85 grams

Putting values in above equation, we get:

$k=\frac{2.303}{6.8min}\log\frac{100}{85}\\\\k=2.39\times 10^{-2}min^{-1}$

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = ?

k = rate constant of the reaction = $2.39\times 10^{-2}min^{-1}$

Putting values in above equation, we get:

$t_{1/2}=\frac{0.693}{2.39\times 10^{-2}min^{-1}}=28.99min$

Hence, the half life of cyclopropane at 760 K is 28.99 minutes.