## 7th edition 7B.13

$\frac{d[R]}{dt}=-k[R]^{2}; \frac{1}{[R]}=kt + \frac{1}{[R]_{0}}; t_{\frac{1}{2}}=\frac{1}{k[R]_{0}}$

Saachi_Kotia_4E
Posts: 68
Joined: Fri Sep 28, 2018 12:23 am

### 7th edition 7B.13

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

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
Posts: 62
Joined: Fri Sep 28, 2018 12:15 am

### Re: 7th edition 7B.13

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

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.