# Apparent contradiction in the type of deviation of an ethanol-water mixture

For a mixture of $\ce{EtOH}$ and $\ce{H2O}$: $\Delta H_{\text{mix}} < 0$ (exothermic) and $\Delta V_{\text{mix}} < 0$ (volume contraction). This means that the unlike interactions are stronger than the like interactions, because the mixture exhibits a lower enthalpy and volume than that would be expected for ideal mixing.

Also, $\ce{EtOH}$ and $\ce{H2O}$ form a minimum boiling azeotrope and the $T-\chi$ graph is below the Raoult's law $T-\chi$ plot. This means that the pair exhibit positive deviation from Raoult's law. This implies that the mixture is more volatile than the pure components. Thus, like interactions should be stronger than unlike interactions.

I am not able to resolve this contradiction. Does the mixture actually exhibit positive deviation? Which interactions are stronger - unlike or like?

• Look for the T dependence of Gmix. It should be positive at about azeotrope boiling point. Commented Apr 11, 2018 at 13:43
• Water-ethanol is a v complicated system. The partial molar volume of ethanol decreases then increases as ethanol mole fraction increases. That of water does the opposite in accordance with Gibbs-Duhem eqn. Entropy changes are also likely to be v important as hydrogen bonding plays such a large part in these mixtures. For example EtOH/water mixtures of the same viscosity but different composition have very different properties. Commented Apr 11, 2018 at 15:40
• But how can it exhibit positive deviation, but still have Vmix and Hmix negative? It is contradictory. Commented Apr 11, 2018 at 17:41
• Those are all T dependent. Check for that as it seems the only explanation. Commented Apr 16, 2018 at 8:40

As suggested in the comments, there is an entropic contribution to the excess free energy which should not be overlooked, as it is the excess Gibbs free energy that determines whether the deviation from Raoult's law is positive or negative. All of the statements made by the OP are correct, except that negative $$\Delta H$$ (exothermicity) and $$\Delta V$$ (volume contraction) do not necessarily imply a reduced volatility (negative deviation from Raoult's law), that is they are not sufficient conditions for a negative deviation, a negative excess Gibbs free energy instead being the sufficient condition. Again, as alluded to in the comments, a simple analysis based entirely on $$\Delta H$$ is not appropriate here, largely because of the hydrogen bonding (associative) nature of the liquids. In fact, the negative mixing volume and enthalpy might partly explain the negative excess entropy (and vice-verse: they are correlated). The excess entropy of mixing of water and ethanol is highly negative over the entire range of composition at room temperature, leading to a net positive excess Gibbs free energy.