# 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. – Alchimista Apr 11 '18 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. – porphyrin Apr 11 '18 at 15:40
• But how can it exhibit positive deviation, but still have Vmix and Hmix negative? It is contradictory. – Renganathan Subramanian Apr 11 '18 at 17:41
• Those are all T dependent. Check for that as it seems the only explanation. – Alchimista Apr 16 '18 at 8:40
• Those are all T dependent. Check for that as it seems the only explanation. – Alchimista Apr 16 '18 at 8:40