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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?

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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.

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  • $\begingroup$ I have an another question too..Why equilibrium constant is proportional to concentration alone??? $\endgroup$ – Archa Sep 26 '18 at 14:13
  • $\begingroup$ @anilbabu What else would you expect it to be proportional to? $\endgroup$ – Aaron Stevens Sep 26 '18 at 14:14
  • $\begingroup$ temperature, pressure,etc $\endgroup$ – Archa Sep 26 '18 at 14:15
  • $\begingroup$ @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. $\endgroup$ – user66472 Sep 26 '18 at 14:44
  • $\begingroup$ This is a first order approximation for ideal dilutions only. If the initial concentration becomes large, $K_c\propto K$ breaks down. $\endgroup$ – Martin - マーチン Sep 27 '18 at 12:56

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