thermodynamics - Kc only effected by temperature?

This makes little sense to me, with a lot of contradictory statements I have learnt.

For example:
$$\ce{N2 + 3H2. <--> 2NH3}$$

($\mathrm{\Delta h = -46.2~kJ~mol^{-1}}$)

I have been told a higher kc = large amount of product being formed in the equilibrium.
So therefore why if I increase the pressure why don't I get a increase in kc as the forward reaction is favoured creating more product?

After asking around I have come to the conclusion that a reaction has many kc values depending on temperature , but then I don't see how the kc equation can work as you can get different concentrations of A B C and have constant temperature by increasing pressure.
And also the statement of higher kc = more product is also wrong.

Obviously I'm wrong but I'm just explaining where my confusion lies in my logic.
Where have I gone wrong? Or is there an explanation that will clear things up?

Answer

We want to perform the reaction you mentioned: $$N_2+3H_2\longrightarrow 2NH_3$$

At the beginning we have certain concentrations of reagents, say $[N_2]_{in}$, $[H_2]_{in}$ and $[NH_3]_{in}$. The reason why I am assuming we have some ammonia is just a way for not dealing with infinite values of energy as it will be clear soon. Since Chemists often work at constant temperature and pressure we will follow the spontaneity of this reaction following the changes in the Gibbs Free energy $G_r(T,P)$ of reaction. A model describing the Gibbs free energy changes is the following: $$\Delta G_{r}(T,P)=\Delta G_r^0(T,P=1bar) + RT \ln\biggl(\frac{[NH_3]^2}{[N_2][H_2]^3}\biggr)$$ The reason we have chosen not to have a zero concentration of ammonia is because in that case we would get $\Delta G_{in}=-\infty$ and this is not sensible. It is important to note that $\Delta G^0$ only depends on the temperature since the pressure has been fixed at 1 bar. Another aspect to note is that the concentrations in square brackets are the concentration at a certain point of the reaction. Now we are ready to find the answer to the question. What happens at the equilibrium? By definition the equilibrium is reached when the Gibbs free energy is minimum, thus when $\Delta G_r =0$. Taking the above equation and setting this condition, isolating the concentrations from the logarithm, we get that the concentrations of our reagents/product at the equilibrium are $$\frac{[NH_3]_{eq}^2}{[N_2]_{eq}[H_2]_{eq}^3}=\exp\biggl(-\frac{\Delta G_r^0}{RT}\biggr)=K_{eq}$$ Clearly the equilibrium constant depends only on the temperature and not on the pressure since in $\Delta G_r^0$ pressure has been fixed at 1 bar by definition. Changing the total pressure does not effect the amount of ammonia you get as you can see from a "mathematical" point of view. However in order to increase the product concentration we can increase the concentration of the reagents (in this case the partial pressure since we are in gas phase) since what is really fixed at the equilibrium for a given temperature is $\frac{[NH_3]_{eq}^2}{[N_2]_{eq}[H_2]_{eq}^3}$ and not the concentrations taken singularly.