Q:

Why is the equilibrium constant of water with its vapour equivalent to its vapour pressure?

Can someone conceptually explain why the equilibrium constant is equivalent to the vapour pressure of water for the above equilibrium reaction?
For example if $K = x$, then it is said that vapour pressure $= x$ (in bars). However I seem to cannot understand why. 

A:

Lets start by writing the definition of the equilibrium constant, for some general reaction:
$$\sum_i^{n_\text{reac}} a_iA_i = \sum_j^{n_\text{prod}}b_jB_j$$
The above is a general reaction, where $a_i$ and $b_j$ is the stoichiometry and $A_i$ and $B_i$ is the molecules. For this reaction we can write the equilibrium constant:
$$K=\frac{\prod_j^{n_\text{prod}}\frac{B_j^{b_j}}{U_j^{\circ}}}{\prod_i^{n_\text{reac}}\frac{A_i^{a_i}}{U_i^{\circ}}}$$
Here $\prod$ is the symbol that denotes we take the product over a range. The $U_i^\circ$ are a standard unit, to ensure that our equilibrium constant is well defined.
Now if we look at your reaction:
$$\mathrm{H_2O\left( l \right)} \rightleftharpoons \mathrm{H_2O\left( g \right)}$$
We can from the above equations identify that; $a_1 = 1$, $b_1=1$, $A_1=\left[ \mathrm{H_2O} \right]$ and $B_1=p_\mathrm{H_2O}$.
We can thus write our equilibrium constant as:
$$K=\frac{\frac{p_\mathrm{H_2O}}{1\text{ bar}}}{\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}}$$
As you might note, I have also inserted the $U$s. For pressure $U=1\text{ bar}$ and for concentrations $U=1\text{ M}$. Now we need one last ingredient. You might have learned that the activity of liquids is $1$, i.e. the means that we have to set $\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}=1$, this is due to the definition of activity. If interested I would suggest to take a look at this. As our final result we can see that:
$$K=\frac{p_\mathrm{H_2O}}{1\text{ bar}}$$
I.e. we have that the equilibrium constant equals the partial pressure of water in standard units.

