7B.13 Help


Moderators: Chem_Mod, Chem_Admin

Xuan Kuang 2L
Posts: 31
Joined: Wed Nov 14, 2018 12:23 am

7B.13 Help

Postby Xuan Kuang 2L » Sun Mar 10, 2019 7:03 pm

Could someone help to explain this question to me?

Q:The half-life of A in a second-order reaction is 50.5s when [A]0= 0.84 mol. L^-1. Calculate the time needed for the concentration of A to decrease to
a) one-sixteenth
b) one-fourth
c) one-fifth

Thank you!

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

Re: 7B.13 Help

Postby Matthew Tran 1H » Sun Mar 10, 2019 7:11 pm

Since the half-life of a second-order reaction depends of [A]0, you can't just multiply the initial half-life by the number of half-lives because you have a different [A]0 for each half-life. Therefore, you have to find k using the equation

Once you have k, you can plug in the the concentrations into the equation and find t.

Philipp_V_Dis1K
Posts: 32
Joined: Fri Sep 28, 2018 12:20 am

Re: 7B.13 Help

Postby Philipp_V_Dis1K » Tue Mar 12, 2019 5:12 pm

1/16= (1/2)^4, so that means 4 times the halflife. 1/4=(1/2)^2 so 2 times the halflife. for part c, use the second order reaction

Matthew Choi 2H
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

Re: 7B.13 Help

Postby Matthew Choi 2H » Wed Mar 13, 2019 5:04 am

For many of these types of problems, I think you should first understand the equations that you would use. Once you know which equations are available, you can maneuver the date given to find the answer you need. For this problem specifically, you need the half life equation and the equation that relates [A]t, [A]0, k and t. However, keep in mind there is a specific order in which you do these problems where you first find information necessary to compute an answer using another equation.

Michael Novelo 4G
Posts: 64
Joined: Fri Sep 28, 2018 12:28 am

Re: 7B.13 Help

Postby Michael Novelo 4G » Wed Mar 13, 2019 11:44 am

Philipp_V_Dis1K wrote:1/16= (1/2)^4, so that means 4 times the halflife. 1/4=(1/2)^2 so 2 times the halflife. for part c, use the second order reaction

this method only applies to first order reactions.


Return to “Second Order Reactions”

Who is online

Users browsing this forum: No registered users and 3 guests