Question

Describe in detail what you expect for the changes in enthalpy, entropy, and free energy when a sample of liquid evaporates. How does temperature affect these changes?

Answer 1

You're pulling the molecules (or, rarely, atoms) of the substance apart, against the attractive intermolecular forces. That will require energy, so the enthalpy change will be positive (you are putting energy in).

Since each molecule will have much more space to move around in the gas phase, compared to the liquid phase, there will me many more microscopic arrangements of the particles consistent with any given temperature and pressure. That means the entropy will have increased. (Entropy is proportional to the number of ways you can arrange the atoms and molecules to give you the same substance at the same temperature and pressure. It's sort of like the number of ways you can combine coins to get the same amount of money: there are only two ways to get $0.06: 6 pennies or 1 penny and 1 nickel. But there are many more ways to get $1.00: 100 pennies, 25 nickels, 10 dimes, 4 quartesr, 3 quarters + 2 dimes + 1 nickel, and so forth. Thus the entropy of $1 in coins is much higher than the entropy of $0.05 in coins.)

The free energy change dG generally is given by a simple equation relating it to the change in enthalpy dH and change in entropy dS:

dG = dH - T dS

where T is the absolutely temperature (measured in kelvins). In this case, since dH > 0 and dS > 0, dG will be positive when T is low (dH dominates) and negative when T is high (dS dominates). There will be a particular temperature, the boiling point, where dG exactly equals zero.