Suppose I have the reversible reaction:

$$\ce{A +B⇌ C} $$

The reaction is at equilibrium with equilibrium constant $K$

I am told that if I increase the concentration of $\ce{B}$, the rate for the forwards reaction will exceed the backwards one. Fair enough.

I am also told that $K$ will necessarily increase. Why though? Its true that Forward reaction > back ward reaction until we reach a new equilibrium such that more of $\ce{C}$ is produced but I don't see why this implies in any way that the final quotient $\frac{[\ce{C}]}{\ce{[A][B]}}$ will necessarily be any greater.

Certainly if we have a simple reaction:

$$\ce{A⇌ C} $$

and we add more of $\ce{A}$, then the equilibrium constant for the new final final state will remain as it was, ceteris paribus.

What am I getting wrong, because my textbook suggests that $K$ will always increase, no matter what type of reaction I am dealing with (of course, as long as all reactants are in a suspended form, e.g. dissolved or gaseous and the reaction is subject to Le Châtelier's principle).

Please, if possible, keep the answer as a level intelligible to a high-school student. I don't know much about advanced university chemistry.

  • 1
    $\begingroup$ K will increase initially as equilibrium is disturbed but it will then return to its original value once a new equilibrium is established (provided the temperature doesn't change). $\endgroup$
    – bon
    Apr 29, 2015 at 20:40
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    $\begingroup$ @bon The equilibrium constant $\rm{K_{eq}=\frac{[C]_{eq}}{[A]_{eq}[B]_{eq}}}$ does not change upon addition or removal of species. The reaction quotient $\rm{Q=\frac{[C]}{[A][B]}}$, however, does change immediately after the equilibrium is disturbed, and with time converges to the same value as $\rm{K_{eq}}$ once more. The distinction is subtle but important, and causes some confusion between students, so it should be made clear. $\endgroup$ Apr 30, 2015 at 1:45
  • $\begingroup$ @NicolauSakerNeto Yes this is true my comment was poorly worded. $\endgroup$
    – bon
    Apr 30, 2015 at 8:38

2 Answers 2


You're correct. The equilibrium constant will not change. The equilibrium constant is only defined at equilibrium. Your analysis of the situation was flawless.


In your book, it was probably mentioned that the Reaction Quotient (Q) will change, K changes only with temperature and there is no other way to change it. Catalysts also effect only the rate (speed) of reaction.


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