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I was watching this video on Khan Academy (about the equilibrium constant), and the analogy that they made really got me thinking. I wanted to verify whether it was completely correct, or just an analogy to explain reaction rates to high school students. I'm well aware that this is an equilibrium constant, and the probability of molecules reacting relates to kinetics. However, this is how it was explained in the video, and I'm trying to make sense of it.

Would anyone please guide me as to whether this analogy, of the molarity and the probability of a molecule reacting, is really true and mathematically correct or whether its just a simplification for students? (I'd recommend that you watch the video first before trying to answer.)

Also, please note that I'm not saying that what Khan Academy is teaching is wrong, I'm just trying to clarify whether what they're teaching is a simplification, or a scientifically accurate explanation.

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  • $\begingroup$ Probability of colliding and reacting is indeed higher as concentration is higher. Note: do not mix thermodynamic (equilibrium of the reaction) with kinetics (how fast equilibrium is achieved) as it seems you perhaps do so comparing title and body of the Q. $\endgroup$ – Alchimista Feb 12 '18 at 13:38
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As Alchimista points out, two different principles are at work here. Concentration, or how many of the species are available, and are these species able to react when they do meet.

The transcript of the video at 3:32 states:

'some constant that takes into account things like temperature and how the molecules are actually configured. Because it's not dependent on them just being there. You have to worry about their kinetic energies...'

So, Molarity will not affect the probability that any two molecules that meet will react, but it does absolutely affect rate of reaction. If the reaction is possible (there is sufficient activation energy) then the more concentrated the substance, the more likelihood of reactable species meeting, and therefore a greater rate of reaction.

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In this video the author is first explaining the origin of the Keq constant based on the rates of the forward and backward reaction and then moves on to explain why the rate law expression is of the form rate=k[A]$^a$[B]$^b$.

No problem with the first part but disagree with the second especially when he says that [A]$^a$[B]$^b$ expresses the probability of having a of A molecules and b of B molecules in a small area at the same time:

Rate = kProbability(Having a of A molecules and b of B molecules in a small area at the same time)

I am no mathematician but the probablity of anything has values between 0 and 1 which is obviously not the case for [A]$^a$[B]$^b$. Only this makes the video wrong.

He also says that:

"The probability of having V molecules in some volume = [Concentration]xvolume"

which is incorrect as well. In a solution of constant concentration you will always have (Conc. x volume) amount of molecules evenly spread in that volume. There is no probability involved. If there is it would be the probability of having (Conc.x volume) molecules of compound in that volume which would be 1 since that is how we define concentration.

In order to understand the form of the rate law equation you should look into collition theory (e.g. http://umich.edu/~elements/03chap/html/collision/index.htm) where you will be able to see exactly how the probability of two coliding molecules reacting or not is properly used.

Note that that the probability of two molecules colliding and reacting is higher with increasing concentration, as Alchimista commented, and this is one of the principles collition theory relies on but not in the way it is presented in the video.

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  • $\begingroup$ I think that the video is getting at 'In order to have a reaction occur, particles need to actually collide, which means be in a region of the overall volume.' So, the greater the concentration, the greater probability of them being in the area of interest. Imagine a 1 mm cube that we can observe in great detail. And I think 'relative probability' is what is meant. $\endgroup$ – MarsJarsGuitars-n-Chars Feb 13 '18 at 23:12
  • $\begingroup$ What you say is correct in qualitatively explaining why increasing the concentration of the reactants of an elementary reaction will lead to an increase in rate. I appreciate the attempt in the video and that it has some parts that are valid, but you can not derive the rate law equation in the way it is described there. It seems like false logic to me and anyway there is already collision theory that does a fine job! $\endgroup$ – AMM Feb 14 '18 at 0:12

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