Question

The isomerization of cyclopropane to form propene is a first-order reaction. At 760 K, 85% of a sample of cyclopropane changes to propene in 79.0 min. Determine the rate constant for this reaction at 760 K.

Answer 1

Answer:

so rate constant  is 4.00 x 10^-4 $s^{-1}$

Explanation:

Given data

first-order reactions

85% of a sample

changes to propene t =  79.0 min

to find out

rate constant

solution

we know that

first order reaction are

ln [A]/[A]0 = -kt

here [A]0 = 1 and (85%) = 0.85 has change to propene

so that [A] = 1 - 0.85 = 0.15.

that why

[A] / [A]0= 0.15 / 1

[A] / [A]0 = 0.15

here t = (79) × (60s/min) = 4740 s  

so

k = - {ln[A]/[A]0} / t

k = -ln 0.15 / 4740

k = 4.00 x 10^-4 $s^{-1}$

so rate constant  is 4.00 x 10^-4 $s^{-1}$