Chemical equilibria and changes in concentration

I learned that if concentration of reactants or products is changed, the position of equilibrium changes but value of $K_c$ remains the same. This just does not make intuitive sense to me... I mean whatever concentration of reactants I start with does not matter, $K_c$ remains same....?

Also, can someone please elaborate the term position of equilibrium

Assuming the following reaction:

$$\ce{A + B <=> C + D}$$

Where:

$$K_\text{c}=1.23\times10^{-2}$$

We say that:

$$K_\text{c}=\frac{\ce{[C]*[D]}}{\ce{[A]*[B]}}=1.23\times10^{-2}$$

And the reaction will be at equilibrium for any concentrations (actually, it should be activities, which are in essence effective concentrations) of A, B, C and D that satisfy the above.

I can understand your confusion. What they mean is... If you have a reaction at equilibrium, and remove half the reactants, then the reaction will proceed towards reactants, until equilibrium is once again reached.

Their way of saying it:

You remove reactants, and the position of equilibrium is shifted towards the left, because the reaction will now proceed towards the left hand side to restore equilibrium.

It's confusing, because equilibrium occurs at the same conditions as before, so you haven't really shifted the equilibrium position in the way it sounds like.

• In general, $K_c=1$ does not imply that you have equal amounts of reactants and products. Consider the reactions $\ce{A + B -> C}$ or $\ce{A + B -> 2C}$ or $\ce{A -> B + C}$. – Nicolau Saker Neto Feb 20 '14 at 18:02
• You're right of course – Brian Feb 20 '14 at 18:15