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


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.

  • $\begingroup$ 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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