1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook