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In my class, it was taught that the equilibrium constant does not vary with the intial concentration of the reactants? Why is it so?

Martin - マーチン
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Archa
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1 Answers1

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Clearly the math tells you that the equilibrium constant is independent of the initial compositions:

For $a\mathrm{A}+b\mathrm{B}\rightarrow c\mathrm{C}+d\mathrm{D}$, $$K_c=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$$ where $[\mathrm{C}]$, $[\mathrm{D}]$, $[\mathrm{A}]$, and $[\mathrm{B}]$ are concentrations at equilibrium.

This makes some sense intuitively: the rate of the reaction is dependent only on the present concentrations of compounds (and some other stuff like pressure and temperature), not previous concentrations. And equilibrium is defined as a position where the rate of forward and backward reactions are the same.

  • I have an another question too..Why equilibrium constant is proportional to concentration alone??? – Archa Sep 26 '18 at 14:13
  • @anilbabu What else would you expect it to be proportional to? –  Sep 26 '18 at 14:14
  • temperature, pressure,etc – Archa Sep 26 '18 at 14:15
  • @anilbabu Exactly. Those are the two other predominant factors. Increasing the pressure moves the equilibrium position towards the side with a smaller number of molecules (assuming all gaseous). Increasing the temperature pushes towards the endothermic direction. Look at Le Chatelier’s principle for some more stuff. –  Sep 26 '18 at 14:44
  • This is a first order approximation for ideal dilutions only. If the initial concentration becomes large, $K_c\propto K$ breaks down. – Martin - マーチン Sep 27 '18 at 12:56
  • @Martin-マーチン Sounds interesting; I just searched online quickly, and I didn't find anything about it (largely because I was unsure what to search for). Do you have any links to stuff about that? –  Sep 27 '18 at 13:03
  • @Martin - マーチン I never knew that $\ce{Kc}$ was an approximation. Can you please elaborate on the actual result then? – Yusuf Hasan Sep 27 '18 at 15:43
  • @YUSUFHASAN The statement that $K_c$ or $K$ doesn't vary with initial concentration is an approximation; and $K_c$ is an approximation to $K$. Maybe my answer here helps What kind of equilibrium constant we use for Gibbs free energy and Van't Hoff equation? (But I have no time to explain this further.) – Martin - マーチン Sep 27 '18 at 15:53