Moderators: Chem_Mod, Chem_Admin

Prasanna Padmanabham 4I
Posts: 111
Joined: Thu Jul 25, 2019 12:17 am


Postby Prasanna Padmanabham 4I » Thu Mar 12, 2020 6:53 pm

Hi Everyone,
I am super stuck on this problem. I would appreciate it if someone could please give me a detailed explanation about how to solve this problem.

Calculate the time required for each of the following second-order reactions to take place: (a) 2A --> B + C, for the concentration of A to decrease from 0.10 mol*L^-1 to 0.080 mol*L^-1, given that kr 5 0.015 L*mol^-1*min^-1 for the rate law expressed in terms of the loss of A;

(b) A --> 2B + C, when [A]0 = 0.15 mol*L^-1, for the concentration of B to increase to 0.19 mol*L^-1, given that kr = 0.0035 L*mol^-1*min^-1 in the rate law for the loss of A.

Thank you!

MingdaH 3B
Posts: 133
Joined: Thu Jul 11, 2019 12:17 am

Re: 7B.17

Postby MingdaH 3B » Thu Mar 12, 2020 6:55 pm

Plug in the values into the equation 1/[A] = kt + 1/[A]0

Anisha Chandra 1K
Posts: 118
Joined: Thu Jul 11, 2019 12:17 am

Re: 7B.17

Postby Anisha Chandra 1K » Fri Mar 13, 2020 8:26 am

For the first part you can just use the numbers for [A] with the above equation. For part b, you would use the fact that delta[A]/delta t = 1/2 delta[B]/delta t. using that you can find out how much [A] changes as [B] increases to 0.19M. You can then use the above equation using [A]0 and [A] when [B] is 0.19M.

Return to “Second Order Reactions”

Who is online

Users browsing this forum: No registered users and 2 guests