When you will read this webpage ,then you will find that they have made cases for different types of second order reaction .

In $situation 2A $ of case 2 in the article (eq 1.13 and 1.14) ,they have taken change in concentration of A to be dx/dt instead of
d(a-x)/dt .What is the reason for this ?

Also what I am also finding difficulty is how did they combine the reactants to make it $[A]^2$....... it would mean that individually the concentration of a single reactant A is following 1st order reaction. So the rate constant in disappearence of A should be first order instead of second so what is the reason for this .I t would have make sense for 2A to follow second order instead of just A

  • $\begingroup$ In linked article all the equations have numbers. Use the numbers to pin point exactly what equations baffle you. Why should we have to guess?!? $\endgroup$ – MaxW Jul 29 '18 at 3:58
  • $\begingroup$ I can't see the equation numbers as they aren't shown in my opened webpage but still I have added more details in my question $\endgroup$ – Scáthach Jul 29 '18 at 5:39

If you are talking about the 1.13 and 1.14 equations, than whta he did was just a simple subistituction where $[A] = x,\ d[A] = dx$, if not, please identify the step.

And, yes, you are going to get diferent equations for $[A]_{(t)}$ depending on the order of reaction, $0^{th}$ order gives you a linear equation, $1^{st}$ order gives you an exponetial equation, and all the orders higher than 1 gives you a fraction diference like: $$\frac{1}{A_t^n}-\frac{1}{A_0^n} = n\ k\ t$$ where n is the order of reaction minus 2.

The step where they make $[A]^2$ is only because you have an elementary equation, it can not be assumed to be that if the equation was golbal or not elementary.


Not the answer you're looking for? Browse other questions tagged or ask your own question.