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I am having a bit of trouble understanding variations to rate equation problems.

What I know:

  1. For $\ce{A + B -> AB}$, if the rate is proportional to the concentration of A and B, the rate equation is:

$\quad$ rate=$\ce{k*[A]^1*[B]^1}$

  1. For $\ce{A + A -> 2B + C}$, if the rate proportional to $\ce{[A]}$:-

$\quad$ rate=$\ce{k*[A]}$ , as per Question 1c and Answer 1c.


Question 1 : Why is this the case?

E.g. when it was $\ce{1A + 1B}$ and the rate was proportional to the concentration of $\ce{A}$ and $\ce{B}$ our rate equation contains $\ce{[A]*[B]}$. But for $\ce{1A + 1A}$ why does the rate equation contain $\ce{[A]}$ instead of $\ce{[A]*[A]}$?

Question 2: Do molecules change anything (compared to single atoms)

Is $\ce{A + A}$ equivalent to $\ce{A2}$

E.g. will all of these have the same rate equation (if the rate is proportional to $\ce{[A]}$)

  • $\ce{A + A -> 2B + C}$
  • $\ce{A2 -> 2B + C}$
  • $\ce{2A -> 2B + C}$

I can't see why they wouldn't all have the same rate equation as it would just be different ways to represent the same total $\ce{[A]}$.

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The issue is that the rate equation is determined not by the reaction equation, but rather by the actual mechanism that lies beneath it. It turns out that reactions that appear fairly simple, such as the burning of methane in oxygen in the flame of a Bunsen burner, do not necessarily take place in a simple manner, I understand that the even the reaction for the Bunsen flame can in real life contain a very large number of steps. Depending on the reaction, different steps will require different reactants, and will take place at different speeds. Maybe for the reaction $\ce{A + B->AB}$, some step that takes a long time involves $\ce{A}$ decomposing on heating, the products of which go on to quickly react further to eventually form $\ce{AB}$. This step that takes a long time is called the rate-determining step, and decides the form of the rate equation. It may be that this causes only $[\ce{A}]$ to feature in the rate equation.

In this case however, the rate equation is given to you by the comments to the right of the equation. Just because we write it as $\ce{2A}$ or $\ce{A+A}$ doesn't make a difference, it's just how someone decided to write the equation, which does not necessarily reflect the steps of the mechanism. In fact, if a mechanism involved a slow step $\ce{A + A -> B}$, the reaction would be second order with respect to $\ce{A}$.

Hopefully that helps.

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