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 $\frac{1}{k}$?