How long does it take for the concentration to drop?

Question

Cyclopropane isomerizes to propene when heated. Rate constants for the reaction,

cyclopropane ---> propene

are 1.10e-4 at 743 K and 1.02e-3 at 783 K

(a) What is the activation energy?

(b) At 773 K and [cyclopropane] = 0.10 M, how long does it take for the concetration to drop to 0.023 M?

Note: i got the first step (a), bu i am still confused with the second step.How can i figure it out.PLS explain step by step.....thanx

Answer 1

The first one you find using the Arrhenius equation:

lnk2/k1 = Ea/R (1/T1-1/T2) to find Ea

for (b) you use the same equation, but now knowing Ea you find k3 for T3=773 K.
Then the reaction is first order, so the remaining concentration is
C=Co*e^-k3t =>
ln(Co/C)= k3t=>
t=ln(Co/C) /k3
Co=0.1
C=0.023
k3 you have calculated

By the way you are not providing units for k1,k2. If it is s^-1 then t will be in s, if it is min^-1 then t in min, etc.