Question

How come higher the value of Henry's Law constant, the lower is the solubility of a gas in a liquid?

Answer 1

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. As `chi_2 -> 0`, we also have `P_2 -> k_(H,chi)chi_2`.

We can see that for a specific solute, we will have a vapor pressure `P_2` specific to that solute's identity at a particular temperature (water vapor can't have a vapor pressure identical to sulfur dioxide if both are at the same temperature, for instance, because water vapor isn't sulfur dioxide).

That is, for as long as we have the same solute at the same temperature, `P_2` will not change.

Since the mole fraction `chi_2` is how much solute has dissolved in the solution, as `chi_2 -> 0`, we can see that `a_(2chi) -> chi_2`, meaning that `a_(2chi) -> 0` as well. Therefore, as solubility decreases, `k_(H,chi)` increases.