2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The equation is pretty much a definition of the equilibrium between liquid water and water vapour. That is what creates a vapour pressure. So, of course, the vapour pressure is tightly related to the equilibrium constant. $\endgroup$ – matt_black Jun 6 '18 at 13:55
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hi Erik, why is $ B_1 = [P_{H_2 O}] $? $\endgroup$ – Dogukan Kayhan Jun 10 '18 at 13:20
  • $\begingroup$ It is not $\left[ p_\text{H_2O} \right]$, but $ p_\text{H_2O}$. I have now corrected this in the answer. Sorry for the confusion. It is the pressure instead of the concentration because it is on gas form. Often when things are on gas form, it is the partial pressure that is interesting instead of the concentration (Concentration would be a very small number). $\endgroup$ – Erik Kjellgren Jun 10 '18 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.