# 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).

• So if I wanted the rate constant of the reaction from the gradient of the graph (explained at the bottom of my question) which equation should I use? Feb 1, 2015 at 15:02
• I don't understand that part of the question. Explain more exactly what you are plotting. Product concentration vs. time? Reactant concentration versus time? Feb 1, 2015 at 15:09
• $t_{1/2}$ against $1/[A]_0$ Feb 1, 2015 at 15:10
• both ways would be equally correct, you just have to explain how you are defining the rate constant. Feb 1, 2015 at 15:19
• How can it be correct to get two different rate constants for one reaction? Feb 1, 2015 at 15:27

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)?

• I have a factor of minus one half outside the integral from the rate equation Feb 1, 2015 at 14:55
• Where did your -1/2 actually come from? Feb 1, 2015 at 15:00
• Stoichiometry of the reaction Feb 1, 2015 at 15:01
• Ah, DavePhD beat me to it, but basically your value is fixed anyway so half of it is the same thing so it's not 2kt. Feb 1, 2015 at 15:02