Henry law and equilibrium
That's an appropriate interpretation of Henry's law, although as you've written it $K$ would be referred to strictly as the solubility constant $H$, see this link to the wikipedia page provided in the comments.

The following shows more formally how to derive the connection between Henry's law, an equilibrium constant and the standard Gibbs free energy for evaporation of the solute.

At equilibrium between the solute in the gas and solution phases, the chemical potential of the solute is equal in both phases, so that 

$$\begin{align} \mu(sol)&amp;=\mu(g)\end{align}$$

In the gas phase (assuming ideal behavior)

$$\begin{align} \mu(g)&amp;=\mu^\circ(g)+RT\log\left(\frac{p_i}{p^\circ}\right)\end{align}$$

Now according to Henry's law, the partial pressure and concentration in the solution are related as 

$$p_i=K\frac{m_i}{m^\circ}$$

(of course, as you point out, this can already be recast as an equilibrium constant expression, but read on)

so that

$$\begin{align} \mu(sol)&amp;=\mu^\circ(g)+RT\log\left(\frac{K(m_i/m^\circ)}{p^\circ }\right)\end{align}$$

Defining 

$$\begin{align} \mu^\circ(sol)&amp;=\mu^\circ(g)+RT\log\left(\frac{K}{p^\circ }\right)\end{align}$$

we can see that 

$$\Delta G^\circ_m = \mu^\circ(g)-\mu^\circ(sol)=-RT\log\left(\frac{K}{p^\circ }\right)=-RT\log\left(K_{eq}\right)$$

which is the familiar relation between the standard Gibbs free energy change and an associated equilibrium constant for the equilibrium $\ce{solute &lt;=&gt; gas}$ provided we define the equilibrium constant as

$$K_{eq} = \frac{K}{p^\circ } = \frac{p_i/p^\circ}{m_i/m^\circ }$$