Moderators: Chem_Mod, Chem_Admin

Amy Xiao 1I
Posts: 101
Joined: Sat Jul 20, 2019 12:15 am


Postby Amy Xiao 1I » Sun Mar 08, 2020 10:34 pm

Dinitrogen pentoxide, N2O5, decomposes by first-order kinetics with a rate constant of 3.7 x 10^-5 /s at 298 K. (a) What is the half-life (in hours) of N2O5 at 298 K?

I am somewhat confused by how the units cancel. Why are the seconds applied to ln2 (numerator) and not the 3.7x10^-5 (denominator)?
Screen Shot 2020-03-08 at 10.32.51 PM.png

Niharika 1H
Posts: 50
Joined: Thu Jul 25, 2019 12:16 am

Re: 7B5a

Postby Niharika 1H » Sun Mar 08, 2020 10:36 pm

0.693/(3.7*10^-5) = 18729.73 s
To get this in hours, we have to divide it by 3600s, to get 5.2 hours.

Return to “General Rate Laws”

Who is online

Users browsing this forum: No registered users and 1 guest