7th edition 7B.13


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Saachi_Kotia_4E
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Joined: Fri Sep 28, 2018 12:23 am

7th edition 7B.13

Postby Saachi_Kotia_4E » 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?

Anusha 1H
Posts: 65
Joined: Fri Sep 28, 2018 12:15 am

Re: 7th edition 7B.13

Postby Anusha 1H » 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)

Andrew Bennecke
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Re: 7th edition 7B.13

Postby Andrew Bennecke » 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.

taywebb
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

Re: 7th edition 7B.13

Postby taywebb » 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.


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