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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?

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    $\begingroup$ Look for the T dependence of Gmix. It should be positive at about azeotrope boiling point. $\endgroup$ – Alchimista Apr 11 '18 at 13:43
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    $\begingroup$ 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. $\endgroup$ – porphyrin Apr 11 '18 at 15:40
  • $\begingroup$ But how can it exhibit positive deviation, but still have Vmix and Hmix negative? It is contradictory. $\endgroup$ – Renganathan Subramanian Apr 11 '18 at 17:41
  • $\begingroup$ Those are all T dependent. Check for that as it seems the only explanation. $\endgroup$ – Alchimista Apr 16 '18 at 8:40
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I know I'm really late to the party, but I was interested in this as well. Essentially, it appears that water and ethanol form a solution that has some structure to it, not unlike micellar aggregation. The OH end of the ethanol molecules aggregate with each other and the polar water, which creates a packing structure that has negative excess volume. However, these aggregated structures still have a bunch of hydrophobic tails sticking out that interact unfavorably with water, resulting in the increased fugacity (effective vapor pressure) associated with positive deviation from Raoult's Law.

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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.

Does the mixture actually exhibit positive deviation? Which interactions are stronger - unlike or like?

Therefore the mixture does exhibit a positive deviation from Raoult's law, even as interaction enthalpies between unlike molecules are stronger.

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