Exothermic mixing of liquids

  1. Why is the mixing of ethanol and water exothermic? Because unless there's a reaction I'm thinking by the law of conservation of energy the total energy of the bonds should remain the same before and after. Because considering hydrogen bonds, the total number of bonds is dependent on the 'limiting' number of donors or acceptors right?

    So since the limiting doesn't change there can't be more bonds being formed right?

    Even if we say that the strength of each H bond changes, wouldn't the strength add up to the same amount again since if some bonds are weaker, others would be stronger?

    Thanks for the help :)
  2. jcsd
  3. I'll try to explain;
    Water molecules held together via H-bonds. Pure ethanol is similar, but the clusters of ethanol are smaller. Of course these two solvents are completely miscible.

    If these two solvents are mixed, the intermolecular bonding structure changes.
    Some water-water H-bonds and ethanol-ethanol H-bonds are broken and some water-ethanol hydrogen bonds are formed.

    The process of breaking bonds is endothermic, and the process of making bonds is exothermic.

    Because molecules are on the inside of the sphere are involved in more H-bonding than
    those on the surface of the sphere, an exothermic mixing dominates.
  4. Hi thanks for the excellent reply :)

    But actually, since the total number of H bonds donors and acceptors remain the same shouldn't the total number of H bonds before and after remain the same?

    Here's my misconception: A bottle of pure water has a total of 100 H bonds and a separate bottle of ethanol has a total 100 H bonds. When I mix them, the total number of H bonds can only reach 200 H bonds. So shouldn't the mixing not give out energy?

    Thanks :)
  5. Borek

    Staff: Mentor

    H-bonds between different molecules aren't equal.
  6. To answer this question in greater detail you need statistical mechanics and water models. The simple answer is that the entropy favors ethanol solvation.
  7. That's true, but shouldn't it add up to zero? Say individually A has stronger H bonds than B individually. So compared to A, A-B's H bonds are weaker but compared to B A-B's H bonds are stronger. So shouldn't they add up to zero?
  8. Borek

    Staff: Mentor

    That would be the case only if the A-B bond energy is an average of A-A and B-B energies, check the math. It doesn't have to.
  9. Oh what do you mean by it doesn't have to? Cos by taking the average it's like if I AA bond energy is 100 and BB 50 then AB would be 75? In this case the total bond strength initially in the
    Two components individually would add up to the same amount as the total bond strength in the mixture.

    Or is it because AB might not be 75? So if it say 100 then more more energy is given out? Then if AB is say 60 then it would mean that energy is absorbed?

    But why does this happen? I was thinking that conservation of energy applied here so the total bond energies should remain the same.
    Last edited: May 19, 2013
  10. Borek

    Staff: Mentor

    AB doesn't have be 75. And conservation of energy is exactly the reason why the mixture changes temperature if AB is not 75.
  11. Ohh that clear things up! Actually linking this to Henry's law for a positive deviation the attractive forces in A and B individually must be greater than between AB. So would that mean if A has 100 and B has 50, AB would have to be say 25? So in this case it would be easier to vapourize it?

    Thanks :)
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