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$$\begin{align}\ce{2A + 2B &-> 2C }\\ \ce{A + B &-> C}\end{align}$$

Why does the equilibrium constant change?

And why the rate of the first reaction square the rate of the second? I understand it mathematically, but logically and scientifically I can't understand it at all.

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  • $\begingroup$ A simple(?) thought experiment: A+B involves a 2-body interaction (i.e. 2 molecules are interacting at one common location). 2A+2B involves a 4-body interaction. That's (MUCH) less than half as likely to occur than 2x 2-body interactions. So there's not much reason to believe that the distribution of reactants and products (i.e. the equilibrium constant) will be in a 1:2 ratio because you doubled the number the particles involved in the reaction, right? $\endgroup$
    – rch
    Apr 22, 2014 at 5:44
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    $\begingroup$ If you consider collisions at hard sphere model, even three bodies hitting each other is highly unlikely. Four however is virtually almost impossible. So the basic question is more likely to be, if there are two of A and B necessary. Upscaling reactions and therefore the change in the kinetics is most likely connected to concentrations. $\endgroup$ Apr 22, 2014 at 6:31
  • $\begingroup$ Thanks @Martin I had meant to make those changes earlier. :) $\endgroup$
    – jonsca
    Apr 22, 2014 at 7:48
  • $\begingroup$ Related question: chemistry.stackexchange.com/questions/6641/… $\endgroup$
    – mic
    Dec 19, 2018 at 2:36

1 Answer 1

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Rate will not be square in most of the cases and who said so? The equilibrium constant scientifically changes because Gibb's free energy is extensive quantity and it is related to equilibrium constant through relation that you might know $\Delta G=\Delta G^\circ +\mathcal{R}T\ln Q $. Where $Q$ is reaction quotient which at equilibrium becomes equilibrium constant and then it gives $\Delta G^\circ = -\mathcal{R}T\ln K$

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  • $\begingroup$ I don't know what Gibb's free energy energy is, what I mean by my question is what happens to the equilibrium constant on doubling the stoichiometry of the equation at constant concentrations??? $\endgroup$
    – Amanda
    Apr 24, 2014 at 8:54
  • $\begingroup$ It will become squared of earlier.Rate doesn't but equilibrium constant does become squared unless all substances are either pure solid or liquid. $\endgroup$ Apr 24, 2014 at 13:47

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