Everybody knows that detergents, or generally amphiphilic substances, lower the surface tension. Of water, usually.

I wonder if that's true for any polar solvent (most likely), i.e. will the same detergents work in, say, ethanol or liquid ammonia?

What happens in a nonpolar solvent? The detergents aggregate on the surface there too, but now with the polar end sticking out. Does the surface tension go up, or down?

And what exactly is the reason for the effect on water anyway, quantitatively? Imagine a surface of water which is covered by exactly a single layer of surfactant molecules, e.g. in a Langmuir trough. Now go to half or double that amount. Is there a discontinuity in the change of surface tension? If yes or no, why?

(sorry for the jumble of questions, I'm not quite sure from which angle to tackle this. The effects seems quite obvious, but I'm not sure it is. My real question is how to explain the effect on a molecular level.)


2 Answers 2


We usually say that at an interface, molecules will orient themselves at a "preferred" position, but we usually ignore that such a "preferred position" can actually be their "least unfavorable" position.

Energetically speaking, the interaction of a lipid chain with a gas is less unfavorable than interaction of a polar group with gas, because interactions from a lipid chain is mainly based on van der Waals interactions which are quite weak compared to polar (i.e. electromagnetic) interactions with a polar group. Also, they cannot make stabilizing bonds with the medium, i.e. water usually.

It must also be noted that the surface tension is linked to the interactions of molecules at the interface.

So "why do surfactants lower the surface tension"?

If you add some "surfactant", aka "tensioactive" agents, in water, its "least unfavorable" position would be at the interface between water and the other medium (usually gas). It will bring down the cohesive energy at the surface, compared to water alone, because it relies on van der Walls interaction. And that is why it lowers surface tension.

  • $\begingroup$ OK, so what's you prediction for soap on an oil drop? Surface tension up or down? I say up, because the polar ends have stronger interaction, so their cohesive energy is higher, and it's harder to stretch them over a larger surface. $\endgroup$
    – Karl
    Mar 28, 2019 at 17:55

Here is the Answer from my friend's Ph.D. Thesis..hope it will help.

When two immiscible liquid phases are in contact with each other the molecules at the interface experience an imbalance of various adhesive and cohesive forces. This leads to an accumulation of free energy at the interface. This excess energy is called surface free energy and can be quantified by a measurement of the energy required to increase the surface area of the interface by a unit amount. This surface energy per unit area can be alternatively described as “interfacial tension” (IFT), which is a force/length measurement. This force tends to minimize the area of the surface, thus explaining why for example liquid drops and air bubbles are round. The common units for interfacial tension (IFT) are dynes/cm or mN/m. [1] A surfactant is a substance that, when present at low concentration in a system, has the property of adsorbing onto the interfaces of the system and of altering to a marked degree the interfacial free energy. A typical liquid-liquid immiscible system comprises of a polar (ex. aqueous) and non-polar (ex. organic oil) component. Surfactants have a characteristic molecular structure consisting of a structural group that has strong attraction for the polar solvent, known as a hydrophilic group, together with a group that has a strong attraction for the non-polar solvent, called the lyophilic group. This is known as an

amphipathic structure. When dissolved in a two-phase liquid-liquid system such as oil-water, the surfactant reorients itself by adsorbing to the oil-water interface, such that its hydrophilic group interacts with the water phase while its lyophilic group interacts with the oil phase. This results in a net decrease in interfacial free energy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.