Phase Changes

A phase change is when matter changes in form, but retains its chemical identity. For example, ice (solid water) will melt to form liquid water, and if heat is added liquid water will boil to form water vapor. The particles of a solid are held in an ordered arrangement. The particles are not able to move although they can vibrate at fixed points. The kinetic energy of the solid particles is negligible, but the interparticle attractions are very strong. As for the liquid state, the particles are able to move with more freedom of motion and the kinetic energy is higher than in the solid state. The interparticle attractions, although strong, are not as strong as when particles are in the solid phase. For the gas phase, particles have a tremendous amount of kinetic energy, and the attractions between particles are negligible.

Melting, Vaporization, and sublimation are endothermic processes. Thermal energy (heat) must be absorbed in order for melting, vaporization, and sublimation to take place. The value of ΔH is positive since the system is absorbing energy. Condensation, solidification (freezing in the case of water), and deposition are exothermic processes. Heat is released from the system and ΔH is negative.

The equations below are for the melting (fusion) of ice, vaporization of water, condensation of water vapor, and for freezing of water (solidification). Note, if a reaction has a positive enthalpy or entropy in one direction, the reverse direction will have the same magnitudes but the sign is changed.

H2O (s) → H2O (l)        ΔHfus = 6.02 kJ/mol;    ΔS = 22 J/(mol⋅K)
H2O (l) → H2O (g)        ΔHvap = 40.7 kJ/mol;    ΔS = 109 J/(mol⋅K)
H2O (g) → H2O (l)        ΔHfus = -6.02 kJ/mol;    ΔS = -22 J/(mol⋅K)
H2O (l) → H2O (s)        Hfus = -40.7 kJ/mol;    ΔS = -109 J/(mol⋅K)

 

The vaporization and melting of H2O are both spontaneous at higher temperatures, ΔG < 0, but nonspontaneous at lower temperatures. For water, vaporization and melting are spontaneous at temperatures greater than 0.00 °C. Solidification (freezing) and condensation are spontaneous at lower temperatures. For water, that would be below 0.00 °C. The heat of sublimation, ΔHsublimation is equal to the sum of the heat of fusion and heat of vaporization.

ΔHsublimation = Hfus + ΔHvap

 

Recall, that the Gibbs free energy is dependent on both enthalpy and entropy.

ΔG = ΔH – TΔS

The Gibbs free energy change determines if a process or reaction is spontaneous. If ΔG is less than zero, the reaction or process is spontaneous at a given temperature. If ΔG is greater than zero, the process or reaction is non-spontaneous at a given temperature. If ΔG is equal to zero, then the process or reaction is at equilibrium.

If we set ΔG to zero, we have

0 = ΔH – TΔS

Rearranging the equation, we get

\(\displaystyle T\;=\;\frac{ΔH}{ΔS}\)

The equation above can be used to determine either a boiling point, freezing or melting point. At the melting point, the solid and liquid are at equilibrium. At the boiling point, both liquid and gas are at equilibrium. The equation can also be used to calculate ΔS, if ΔH and the temperature are known, or ΔH, if ΔS and the temperature are known. For example, benzene has ΔHvap = 30.7 kJ/mol and ΔSvap = 87.0 J/(mol#8901;K). We can find the boiling point of benzene.

\(\displaystyle T\;=\;\frac{30.7\;kJ/mol}{0.0870\;kJ/(mol⋅K)}\;=\;352.9\;K\;=\;79.0°C\)

 

Note, the heat of vaporization is always greater than the heat of fusion. This is because the energy is being put into completely disrupting the particle attractions during vaporization while during melting, heat is being put into increasing the distance between particles. It will take much more energy to completely disrupt the particle attractions.

Worksheet: Phase Changes

Exercises

Exercise 1. Methanol, CH3OH, has a boiling point of 64.7 °C and a standard enthalpy of vaporization of 37.4 kJ/mol. What is the standard entropy of vaporization in J/(mol⋅K)?

Check Answer/Solution to Exercise 1

Exercise 2. Pentane, C5H12, has an enthalpy of fusion equal to 8.4 kJ/mol and an entropy of fusion equal to 58.5 J/(mol⋅K). What is the melting point of pentane?

Check Answer/Solution to Exercise 2

Exercise 3. Explain why ΔHvap is always greater than ΔHfus.

Check Answer/Solution to Exercise 3

Exercise 4. For the following processes indicate the sign of ΔH and ΔS. Indicate the type of process.

a) NH3 (g) → NH3 (l)
b) CO2 (s) → CO2 (g)

Check Answer/Solution to Exercise 4

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