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Question regarding hydrophobic forces

  1. Aug 4, 2007 #1
    I've been reading a bit, and I thought I understood it, but now I'm really confused:

    I know that hydrophobic forces at room temperature are almost entirely entropy driven: because a system aims for maximal entropy, it pushes non-polar molecules into an aggregate in order to reduce the clathrate structures that water must form.

    I also know that if you increase the temperature, then the hydrophobic forces get stronger.

    Here's the problem: dG = dH - TdS;
    At equilibrium, dG = 0 so dH = TdS.
    In other words, if I increase the temperature, the entropy decreases. This would mean that a decrease in entropy leads to an increase in hydrophobic forces. Isn't this contradictory? I mean, a decrease in entropy would mean more order = more clathrate structures, so shouldn't this result in a decrease in hydrophobic forces? What's wrong with my reasoning?
  2. jcsd
  3. Aug 5, 2007 #2

    I am not an expert on thermodynamics but does the enthalpy not increase as well with temperature - the values you get from a table is at 298 K so ....

    By just raising the temperature your system is no longer in equilibrium this might be an approximation for certain temperature ranges - the enthalpy is a function of the heatcapacity and temperature changes. At equilibrium you can than see if the entropy of your system has increased.
  4. Aug 6, 2007 #3
    For an event to be spontaneous G has to be negative so if entropy is increased the TdS term becomes larger, likewise with temperature, so G is more negative.

    As stated above, changing T shifts the eqm position so dH (at the old T) is no longer = to TdS (at the new T).
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