64.6k views
3 votes
The reaction ch3nc -> ch3cn is first order. it takes 156 seconds for the concentration of reactant to fall from 0.100 m to 0.0500 m. how much time would it take for the concentration of reactant to fall from 0.0500 m to 0.0400 m?

1 Answer

4 votes
Reaction of interest is: CH3NC → CH3CN

Given: Reaction obeys 1st order kinetics

Also, it takes 156 s for the concentration of reactant to fall from 0.100 m to 0.0500 m. Hence, half life of reaction (t1/2) is 156 s.

We know that, for 1st order reaction:
k = 0.693 / t(1/2)
= 0.693 / 156
= 0.0044 s-1

We also know that, for 1st order reaction
t = 2.303/k log (initial conc./ final conc.)

∴ Time required for the conc. of reactant to fall from 0.05 m to 0.04 m = 2.303/k log (initial conc./ final conc.)
= 2.303/0.0044 log(0.05/0.04)
= 50.72 s.

User Difurious
by
5.5k points