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:


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:


some write:


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

  • $\begingroup$ 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? $\endgroup$
    – RobChem
    Feb 1 '15 at 15:02
  • $\begingroup$ I don't understand that part of the question. Explain more exactly what you are plotting. Product concentration vs. time? Reactant concentration versus time? $\endgroup$
    – DavePhD
    Feb 1 '15 at 15:09
  • $\begingroup$ $t_{1/2}$ against $1/[A]_0$ $\endgroup$
    – RobChem
    Feb 1 '15 at 15:10
  • $\begingroup$ both ways would be equally correct, you just have to explain how you are defining the rate constant. $\endgroup$
    – DavePhD
    Feb 1 '15 at 15:19
  • $\begingroup$ How can it be correct to get two different rate constants for one reaction? $\endgroup$
    – RobChem
    Feb 1 '15 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)?

  • $\begingroup$ I have a factor of minus one half outside the integral from the rate equation $\endgroup$
    – RobChem
    Feb 1 '15 at 14:55
  • $\begingroup$ Where did your -1/2 actually come from? $\endgroup$
    – Talisman
    Feb 1 '15 at 15:00
  • $\begingroup$ Stoichiometry of the reaction $\endgroup$
    – RobChem
    Feb 1 '15 at 15:01
  • $\begingroup$ 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. $\endgroup$
    – Talisman
    Feb 1 '15 at 15:02

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