I know that Raoult's law can be a special case of Henry's law when the value of the Henry's constant becomes equal to the vapour pressure of the pure state of the liquid. Why can't we say the opposite, i.e. Henry's law is a special case of Raoult's law?
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Raoult's law is dealing with the equilibrium between the gas phase and a liquid phase of two liquids. Henry's law is dealing with a gas phase in equilibrium with gas dissolved in a liquid. I don't see how either is a special case of the other. – MaxW May 05 '20 at 08:34
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If we formally consider the Raoult scenario as gas(vapour ) dissolving in liquid, then these mixed solvents are special case of more general solutions the Henry law applies to. – Poutnik May 05 '20 at 09:27
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@MaxW That's what confused me, because they seem unrelated to me. But it's an actual subsection in my textbook, where the both are compared and Raoult's law is considered a special case if the above mentioned criteria gets satisfied. – Nandini May 05 '20 at 09:42
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@Poutnik Can you please clarify what kind of general solutions would Henry's law apply to? – Nandini May 05 '20 at 09:49
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@Nandini Rather all solutions where Henry law is valid. – Poutnik May 05 '20 at 09:51
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@Poutnik - Thanks for the correction. – MaxW May 05 '20 at 17:19
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Related: What are the key differences between Raoult’s Law and Henry’s Law – ananta May 22 '23 at 05:13
1 Answers
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Raoult's law is a proposition regarding the relationship between the vapor pressure of a solvent (component 1 below) in its pure form ($p_1^\circ$) and when a solute (component 2) is present at a mole fraction $\chi_2=1-\chi_1$ (the vapor pressure then $p_1$):
$$\chi_1 = \frac{p_1}{p_1^\circ} $$
It is generally true for very dilute solutions because we know independently that
$$\frac{p}{p^\circ} \rightarrow1 \,\, \text{as} \,\,\chi_1 \rightarrow 1$$
For some ideal binary systems Raoult's law is applicable over a broad range of composition. In fact, sometimes the solute also observes Raoult's law in dilute solutions:
$$\chi_2 = \frac{p_2}{p_2^\circ} $$
This however is exceptional behavior. It is more common that the solute observes the general Henry's law
$$\chi_2 = \frac{p_2}{K_H} $$
where $K_H$ is a proportionality constant for the solute-solvent pair. It should be evident that when $K_H=p_2^\circ$ the solute displays Raoult's law behavior even in dilute solution.
Henry's law is a general statement observed by solutes with a measurable vapor pressure $p_2$, since the following limiting behavior is expected as solute-solvent interactions far exceed solute-solute interactions in importance in the limit of zero solute concentration:
$$\frac{p_2}{K_H} \rightarrow1 \,\, \text{as} \,\,\chi_1 \rightarrow 1$$
However, there is no principle that guarantees that $K_H=p_2^\circ$. Again, for a substance to follow Raoult's law at low concentrations is regarded as exceptionally ideal behavior.
Buck Thorn
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Hi, I do not understand why $\frac{p_2}{K_H}\to 1$ as $\chi_1\to 1$. Shouldn't we, by Henry’ Law, have $\frac{p_2}{K_H}\to 0$ because $\chi_2\to 0$? – insipidintegrator Aug 23 '22 at 13:04
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@insipidintegrator $\chi_1\to 1$ and $\chi_2\to 0$ are equivalent for a two component solution since $\chi_1+\chi_2=1$ – Buck Thorn Aug 24 '22 at 06:56
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No wait so we should have $\frac{p_2}{K_H}\to 0$ right? I understand they are equivalent, that’s why I asked. Because $p_2\propto \chi_2$ – insipidintegrator Aug 24 '22 at 06:57