How come higher the value of Henry's Law constant, the lower is the solubility of a gas in a liquid?
1 Answer
We have the equation for Henry's law standard state of an ideal binary mixture as (Physical Chemistry: A Molecular Approach, McQuarrie, Ch. 25-2)
\mathbf(a_(2chi) = P_2/(k_(H,chi))) ,where:
a_(2chi) is the activity of the solute based on a mole fraction scale (chi_2 = (n_2)/(n_1 + n_2) ).As
chi_2 -> 0 ,a_(2chi) -> chi_2 , i.e. the smaller the amount of solute, the more appropriate it is to use mole fractions instead of activities to describe the amount of solute in solution.
P_2 is the vapor pressure of the solute in a vapor-liquid solution.k_(H,chi) is the Henry's Law constant on a mole fraction scale. Aschi_2 -> 0 , we also haveP_2 -> k_(H,chi)chi_2 .
We can see that for a specific solute, we will have a vapor pressure
That is, for as long as we have the same solute at the same temperature,
Since the mole fraction