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).


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




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


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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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