While studying about Henry's Law, I encountered this equation,I didn't know how to derive it so I searched about it and got nothing, I tried to substitute $H$ (Henry's const.) by $\gamma\cdot f_i$ where $\gamma$ is activity coefficient at infinite dilution and $f_i$ is fugacity of pure species, but still had no idea what to do to reach a formula that I'm familiar with; so I welcome any idea about how to derive, or how to reverse engineering the formula.

  • $\mathrm{d}\left(\frac1T\right)$ is $-\frac{1}{T^2}\cdot \mathrm dT$ and the $\frac{\Delta H}{RT^2}$ reminds me of some kind of $\Delta G$ equation:

$$\frac{\mathrm d\ln H}{\mathrm d(1/T)}= \frac{-\Delta_\textrm{sol}H }{R}\;.$$


Assuming Henry's coefficient $\mathcal{H}_i$ is defined by the following relation:

$$ \mathcal{H}_i = \gamma_i^\inf \cdot p_i^\mathrm{sat} $$

Where $\gamma_i^\inf$ is the activity coefficient for an infinite diluted solution and $p_i^\mathrm{sat}$ is the vapor pressure of pure component $i$.

One can try to extrapolate Henry's coefficient at another temperature by assessing the following ratio (this is a common approach in Chemistry, not always valid):

$$ \frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\gamma_i^\inf(T_1) \cdot p_i^\mathrm{sat}(T_1)}{\gamma_i^\inf(T_0) \cdot p_i^\mathrm{sat}(T_0)} $$

The RHS maybe rearranged in order to have the well know form of ratio of mixed Equilibrium Constants $K_i$ considering that bulk concentration are close and ratio of activities also simplifies (over a small range of temperature).

$$ \frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\frac{\gamma_i(T_1)\cdot p_i(T_1)}{\eta_i(T_1)\cdot x_i(T_1)}}{\frac{\gamma_i(T_0)\cdot p_i(T_0)}{\eta_i(T_0)\cdot x_i(T_0)}} =\frac{K_i(T_0)}{K_i(T_0)} $$

That is, we are stating that ratio of Henri constants approximates ratio of equilibrium constant because we assume that only ratio of pressure is significant and other ratios tend to unity. This is, off course, not always acceptable, but in some restrained conditions it might hold.

Where the concerned reaction is the converse of gas solubilisation:

$$ \mathrm{X}_{i,(\mathrm{aq})} \rightleftharpoons \mathrm{X}_{i,(\mathrm{g})}\quad \Delta_\mathrm{R}H = -\Delta_\mathrm{sol}H $$

And its mixed Equilibrium Constant:

$$ K_i = \frac{\gamma_i\cdot p_i}{\eta_i \cdot x_i} $$

Then one may apply Van't Hoff relation using Enthalpy $\Delta_\mathrm{sol}H$ of the concerned reaction, from its integrated form, it comes:

$$ \frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \exp\left[\frac{\Delta_\mathrm{sol}H}{R}\left(\frac{1}{T_1}-\frac{1}{T_0}\right)\right] $$

This formulae is generally a good approximation around 20°C over a small range of temperature. Do not forget that $\Delta_\mathrm{sol}H$ is not a constant but rather a function of temperature.

If you wish to get your definition, you just have to differentiate the last relation with respect to temperature:

$$ \frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}T} \approx \frac{\Delta_\mathrm{R}H}{RT^2} = -\frac{\Delta_\mathrm{sol}H}{RT^2} $$

Or equivalently: $$ \frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}\left(\frac{1}{T}\right)} \approx -\frac{\Delta_\mathrm{R}H}{R} = \frac{\Delta_\mathrm{sol}H}{R} $$

  • $\begingroup$ Can you please explain how you changed the right side of second equation? $\endgroup$ – Mo Samani May 24 '16 at 16:15
  • $\begingroup$ assuming that ratio of $x's$ and ratio of activity coefficients are 1 right? so what's the formula used for $K$ here ? $\endgroup$ – Mo Samani May 24 '16 at 16:28
  • $\begingroup$ How the equation for $Ki$ has obtained ? I know $K$ is the ratio of $fi$ to $fi$ (standard condition)، I understand the numerator,But Can't understand the denominator. $\endgroup$ – Mo Samani May 25 '16 at 10:17
  • $\begingroup$ This is the expression of a mixed (polyphasic) equilibrium constant. $x_i$ stands for the bulk dissolved gas concentration at equilibrium and $\eta_i$ is its activity coefficient. It is just the ratio of product/reactant as you always write them for aqueous solutions. $\endgroup$ – jlandercy May 25 '16 at 11:35
  • $\begingroup$ If you feel uncomfortable with this mixed constant, try convert it into $K_c$ or $K_p$. You will get some $RT$ terms that simply because of the ration of constants. $\endgroup$ – jlandercy May 25 '16 at 11:44

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