Second order reaction half life equation?

Question

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 $\frac{1}{k}$?

Answer 1

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

Answer 2

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)?