Question regarding hydrophobic forces

[johndoe3344]

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?

 

[greisen]

Hi,

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.

 

[thearny]

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).