One Response to “How is the boiling point of a substance a trade-off between enthalpy and entropy?”

1. hfshaw says:

   June 8, 2010 at 2:14 am

   At the boiling point of a substance, the liquid and gaseous forms of the substance are in equilibrium. That means that for the reaction:

   X_liquid = X_gas

   the free energy change is zero:

   0 = ΔG = ΔH_vap – T_boiling*ΔS_vap

   where ΔH_vap and ΔS_vap are the enthalpy and entropy of vaporization, respectively, and T_boiling is the boiling temperature (all this assume that a specific pressure has been picked, and is held constant).

   Rearranging this, we have:

   ΔH_vap = T_boiling*ΔS_vap

   ΔH_vap/ΔS_vap = T_boiling

   The boiling temperature is therefore given by the ratio of the enthalpy to the entropy of vaporization. If the enthalpy change required to vaporize a material increases, so does the boiling temperature. If the entropy change accompanying vaoprization increases, the boiling temperature decreases.

   Note that ΔH_vap is the energy needed to convert a specified amount of the substance from a liquid to a gas at constant pressure. Because it always takes energy to do this (boiling is always an endothermic reaction), ΔH_vap is always positive.

   T is measured on the thermodynamic scale (in kelvins), and is always positive. Because ΔH_vap is always positive, this means that ΔS_vap is also always positive; the molar entropy of a gas is always larger than the molar entropy of the liquid.