### 7th edition 7B.13

Posted:

**Mon Mar 04, 2019 3:21 pm**how do you find the time it takes for the second order reactions to take place in 7B.13? what formula do we need to use?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=149&t=43594

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Posted: **Mon Mar 04, 2019 3:21 pm**

how do you find the time it takes for the second order reactions to take place in 7B.13? what formula do we need to use?

Posted: **Mon Mar 04, 2019 5:38 pm**

you use the half life equation for second order reactions first. With it, you can use the half life and [A]0 value they give you to solve for k.

once you have k, you can solve for the t values by using the integrated rate law equation

(you'll have to use (1/n)([A]o) where 1/n represents the concentration decrease)

once you have k, you can solve for the t values by using the integrated rate law equation

(you'll have to use (1/n)([A]o) where 1/n represents the concentration decrease)

Posted: **Sun Mar 10, 2019 3:50 pm**

The half-life of A in a second-order reaction is 50.5 s when [A]not=0.9=84 mol*L^-1. Calculate the time needed for the concentration of of A to decrease to (a) one-sixteenth; (b)one-fourth; (c) one-fifth of its original value.

Plugging-in the calculated k value and the given initial concentration into the integrated rate law equation, the only variable left to solve for is t.

Plugging-in the calculated k value and the given initial concentration into the integrated rate law equation, the only variable left to solve for is t.

Posted: **Fri Mar 15, 2019 3:07 pm**

First, you find k using the second order half-life equation, then you isolate t in the integrated rate law to solve for it. Because we already have the ratio of initial to final concentrations, you can use that to relate the two fractions together, making [A]t the same as [A]0. You can then subtract using the ratio and plug in the k you solved for and the given initial concentration.