Quiz Prep Question

Moderators: Chem_Mod, Chem_Admin

Kira Conde 2O
Posts: 24
Joined: Fri Jul 15, 2016 3:00 am

Quiz Prep Question

Postby Kira Conde 2O » Mon Feb 20, 2017 10:39 pm


I found this question in an old quiz prep book and I'm not sure how to do it. It reads:

When 420 mg of NO2 is confined to a 150 ml reaction vessel and heated to 300 C it decomposes by a second order process with the rate of k= 0.54 l/mol/s. What is its initial rate? By what factor will the rate increase if the initial concentration is changed by 1.68 grams?

I honestly don't even know where to start! Please help!

Jessica Huang 1M
Posts: 26
Joined: Wed Sep 21, 2016 2:59 pm

Re: Quiz Prep Question

Postby Jessica Huang 1M » Mon Feb 20, 2017 11:54 pm

second order process means that rate=k[NO2]^2

must find [NO2]: 0.42 g NO2*(mol NO2/46 g)/0.15 L = 0.06087 mol NO2/L

rate= k[NO2]^2
=(0.54 L/mol/s)* (0.06087 mol NO2/L)^2 = 0.0020 mol/l/s = initial rate

For second part of the question, repeat above process with new initial amount of NO2 and then compare your new rate with your old rate.

Return to “General Rate Laws”

Who is online

Users browsing this forum: No registered users and 1 guest