Second order reaction half life equation?

For the reaction: 
$$2A\to p$$
The rate, $v$ may be given as:
$$v=-\frac12\frac{d[A]}{dt}=k[A]^2$$
Correct?


Integrating this gives:
$$-\frac12\int\_{[A]\_0}^{[A]}\frac{d[A]}{[A]^2}=k\int\_0^tdt$$
$$\frac1{2[A]}-\frac1{2[A]\_0}=kt$$
Rearranging...
$$\frac1{[A]}=\frac1{[A]\_0}+2kt$$
Now substituting $t=t\_{1/2}$ and $[A]=\frac{[A]\_0}2$...
$$\frac2{[A]\_0}=\frac1{[A]\_0}+2kt\_{1/2}$$
Minus $1/[A]\_0$ from each side...
$$\frac1{[A]\_0}=2kt\_{1/2}$$
This then gives:
$$t\_{1/2}=\frac1{2k[A]\_0}$$


However, I have seen in textbooks that this should be written:


$$t\_{1/2}=\frac1{k[A]\_0}$$


Have I gone wrong somewhere? If so where?


I need to know which equation is correct because when figuring out the rate constant, would the gradient be equal to $\frac{1}{2k}$ or $\frac{1}{k}$?




What you did isn't wrong, but instead of:


$$v=-\frac12\frac{d[A]}{dt}=k[A]^2$$


some write:


$$v=-\frac{d[A]}{dt}=k[A]^2$$


which is also true.


Twice a constant is still as constant (a different constant of course).


See <http://www.chem.arizona.edu/~salzmanr/480a/480ants/int2ord/int2ord.html>


 6

This is purely a mathematical answer, but shouldn't the integration produce 


1/[A] - 1/[A]0 = kt


which means you don't have your 2kt being taken through your working, and gives you the same as the textbooks' answer(s)?


 0