# Equilibrium constant when adding more of a reactant

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).

• 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).
– bon
Apr 29 '15 at 20:40
• @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. Apr 30 '15 at 1:45
• @NicolauSakerNeto Yes this is true my comment was poorly worded.
– bon
Apr 30 '15 at 8:38