## 14.33 Equation for the half-life of a 2nd order reaction

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

Hannah Pablo 2H
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Joined: Fri Sep 26, 2014 2:02 pm

### 14.33 Equation for the half-life of a 2nd order reaction

14.33 The half-life for the second-order reaction of a substance A is 50.5 s when [A]0 =0.84 mol/L. Calculate the time needed for the concentration of A to decrease to (a) one-sixteenth; (b) one- fourth; (c) one-fifth of its original value.

How do you obtain the equation to begin this problem?

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 14.33 Equation for the half-life of a 2nd order reaction

Half life for second order
At the half life, t1/2, half the initial concentration remains. Substituting this piece of information into the second order integrated rate law and solving for t gives the equation for the half life of a second order reaction.
Last edited by Niharika Reddy 1D on Sun Feb 22, 2015 7:21 pm, edited 1 time in total.

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 14.33 Equation for the half-life of a 2nd order reaction

On Prof. Lavelle's Constants and Equations, the last line gives the two relevant equations for this problem.
For second order reactions, use $\frac{1}{[A]} = kt + \frac{1}{[A]}_{0}$ and $t_{1/2} = \frac{1}{k[A]_{0}}$ (the sheet uses R instead of A, but I used A because that's what the solutions manual uses).

Rearrange the half-life equation to solve for k.

Then rearrange the first equation to solve for t. Replace $[A]$ with $\frac{1}{n}[A]_{0}$ where $\frac{1}{n}[A]_{0}$ is the concentration after the initial concentration has decreased to one-nth its original value.