# Equilibrium constant Kp for vaporization of water?

For the reaction H2O(l) -->H2O(g) it is given that Kp = p(H₂O(g))/p° where p° is the atmospheric pressure. Why is it not just equal to p(H₂O(g))?

• It is (numerically, I mean). – Ivan Neretin May 16 '18 at 14:15
• @IvanNeretin But that is only for the case of unit atmospheric pressure right? And this is the first time I'm seeing any mention of atmospheric pressure in equilibrium constant. Is this the case with all equilibrium constants? – Hema May 16 '18 at 14:17
• Let's put it this way: you want your constant to be greater than 1 when and only when the reaction is shifted to the right, don't you? Well, here you go. – Ivan Neretin May 16 '18 at 14:45
• @IvanNeretin would you please explain a bit more, I didn't really understand.. if we didn't put the atmospheric pressure would it be greater than one? And why do we use the atmospheric pressure in this particular reaction when we don't use it in many others? – Hema May 16 '18 at 15:01

## 1 Answer

Recall the origin of the equilibrium constant. It comes from the fact that the chemical potential of an ideal gas can be written in the form u(T,p0) + RTln(p/p0), with p0 some reference pressure. What this lets you do is measure the equilibrium composition at a reference pressure, then calculate the equilibrium composition at a slightly different pressure with a simple formula, without having to know the general dependence of chemical potential on pressure and temperature. Very handy! But the existence of the reference pressure is essential. It goes into the numerical value of the equilibrium constant. For example, for just two reactants with stoichiometric coefficients c1 and -c2, i.e. the equilibrium is c1 R1 = c2 R2, you have:

dG = 0 = c1 (u1 + RTln(p1/p0)) - c2 (u2 + RTln(p2/p0))

-(c1 u1 - c2 u2)/RT = ln((p1/p0)^c1 / (p2/p0)^c2)

exp(-(c1 u1 - c2 u2)/RT) = (p1/p0)^c1 / (p2/p0)^c2

The thing on the left is just a function of T and the reference pressure p0, and we set it equal to the measured equilibrium constant K(T,p0), which of course we get by measuring the equilibrium composition at T and p0. The thing on the right you'll then recognize as the usual expression for Kp.

K(T,p0) = (p1/p0)^c1 / (p2/p0)^c2

With this equation in hand (and the value of K) we can calculate the equilibrium composition {p1,p2} at some temperature and pressure not too far away from the reference T and p0, using the usual freshman chemistry tricks.

Now you can certainly choose a different reference pressure if you like, it doesn't have to be 1atm, that's just convention. If you chose a different pressure, K would be different, but the expression for Kp would be the same, only you'd be dividing by the new reference pressure. But you have to have some reference pressure for this method to work.