Wednesday, Dec. 5th: “A” DayThursday, Dec. 6th: “B” Day

AgendaHomework Questions/Problems?Sec. 11.2 Quiz: “Intermolecular Forces”Section 11.3: “Energy of State Changes”

Enthalpy, Entropy, and Changes of Sate, Gibbs Energy and Changes of State, Calculating Melting and Boiling Points

Lab write-up: “Energy and Entropy: Phase Changes”Homework

Sec 11.3 review, pg. 398: #1-3, 6-9“What Would Life Be Like?” PaperConcept Review: “Energy of State Changes”

Homeworkpg. 392: #6-12

Problems?

Questions?

Section 11.2 Quiz“Intermolecular Forces”

You may use your notes and your book to complete the quiz on your own…

Good Luck!

Enthalpy, Entropy, and Changes of State

Enthalpy is the total energy of a system.

Entropy measures a system’s disorder.

The energy added during melting or removed during freezing is called the enthalpy of fusion.

Particle motion is more random in the liquid state, so as a solid melts, the entropy of its particles increases. This increase is called the entropy of fusion.

Enthalpy, Entropy, and Changes of State

As a liquid evaporates, a lot of energy is needed to separate the particles. This energy is called the enthalpy of vaporization.

Particle motion is much more random in a gas than in a liquid. A substance’s entropy of vaporization is much larger than its entropy of fusion.

Enthalpy and Entropy Changes for Melting and Evaporation

Enthalpy and entropy both change as energy (in the form of heat) is added to a substance.

The energy added as ice melts at its melting point is the molar enthalpy of fusion (∆Hfus).

∆Hfus is the difference in enthalpy between solid and liquid water at 273 K (O˚C).

∆Hfus = H(liquid at melting point) H(solid at melting point)

Enthalpy and Entropy Changes for Melting and Evaporation

Enthalpy and Entropy Changes for Melting and Evaporation

The energy added as liquid evaporates at its boiling point is the molar enthalpy of vaporization (∆Hvap).

∆Hvap is the difference in enthalpy between liquid and gaseous water at 373 K (100˚C).

∆Hvap = H(vapor at boiling point) H(liquid at boiling point)

Enthalpy and Entropy Changes for Melting and Evaporation

Because intermolecular forces are not significant in the gaseous state, most substances have similar values for molar entropy of vaporization, ∆Svap.

Gibbs Energy and State ChangesThe relative values of H and S determine whether

any process, including a state change, will take place.

A change in Gibbs energy is defined as:

∆G = ∆H T∆S

A process is spontaneous if ∆G is negative OR if ∆S is positive.

If ∆G is positive, then a process will not take place unless there is an outside source of energy.

Gibbs Energy and State Changes

If ∆G is zero, then the system is at equilibrium.

At equilibrium, the forward and reverse processes are happening at the same rate.

For example, when solid ice and liquid water are at equilibrium, ice melts at the same rate that water freezes.

Determining Melting and Boiling Points

For a system at the melting point, a solid and a liquid are in equilibrium, so ∆G is zero.

Remember, ∆G = ∆H - T∆S and if ∆G is zero, then ∆H = T∆S.

Rearranging the equation, you get the following equation for the melting point of a substance:

Determining Melting and Boiling Points

For a system at the boiling point, a gas and a liquid are in equilibrium.

Rearranging the equation as before, you get the following equation for boiling point :

When ∆Hvap > T∆Svap, the liquid state is favored.When ∆Hvap < T∆Svap, the gaseous state is favored.

Pressure Can Affect Change-of-State Processes

Boiling points are pressure dependent because pressure has a large effect on the entropy of a gas.

When a gas expands (pressure is decreased) its entropy increases because the degree of disorder of the molecules increases.

Thus, boiling points change as atmospheric pressure changes due to changes in elevation.

Pressure Can Affect Change-of-State Processes

Liquids and solids are almost incompressible.Therefore, changes in atmospheric pressure

have little effect on the entropy of substances in liquid or solid states.

Also, ordinary changes in pressure have essentially no effect on melting and freezing.

Thus, melting and freezing points hardly change at all with changes in elevation.

Sample Problem APg. 397

The enthalpy of fusion of mercury is 11.42 J/g, and the molar entropy of fusion is 9.79 J/K mol. The ∙enthalpy of vaporization at the boiling point is 294.7 J/g, and the molar entropy of vaporization is 93.8 J/K mol. Use the molar mass of mercury, ∙200.59 g/mol to calculate the melting point and the boiling point.

Practice Problem #1Pg. 397

For ethyl alcohol, C2H5OH, the enthalpy of fusion is 108.9 J/g, and the entropy of fusion is 31.6 J/K mol. The enthalpy of vaporization at the ∙boiling point is 837 J/g, and the molar entropy of vaporization is 109.9 J/K mol. Calculate the ∙freezing and boiling points for each substance.

(Hint: you need the molar mass of ethyl alcohol)

“Energy and Entropy: Phase Changes” Lab Discussion

As heat energy flows from a liquid, its temperature drops.

The entropy (degree of disorder) also decreases.The continuous flow of energy from a liquid will

cause the liquid to eventually undergo a phase change to a solid.

“Energy and Entropy: Phase Changes” Lab Discussion

During the time needed for the phase change to be completed, the temperature of a pure substance will not change.

Entropy, however, will continue to decrease.Once the phase change is complete, the

temperature of the solid will decrease as energy continues to be removed.

“Energy and Entropy: Phase Changes” Purpose

In this lab, you will monitor the temperature of sodium thiosulfate pentahydrate (Na2S2O3 5 H∙ 2O) as it is cooled to several degrees below its freezing temperature and then warmed to several degrees above its melting temperature.

The results of the lab will allow you to determine the freezing and melting temperatures of sodium thiosulfate pentahydrate and to interpret changes in energy and entropy.

“Energy and Entropy: Phase Changes” Materials

The materials for this lab are as listed under “Apparatus” and “Materials” section of the “Experiment 5” handout.

You will be using a bunsen burner, NOT a hot plate.

Listen for other changes….

“Energy and Entropy: Phase Changes”

Procedure For this lab, you will follow the procedure

from the “Experiment 5” handout but complete the data table, analysis, conclusion, and extension questions from the “Construction a Heating/Cooling Curve” handout.

HomeworkSection 11.3 review, pg. 398: #1-3, 6-9Finish lab write-up: “Energy and Entropy: Phase

Changes” Concept Review: “Energy of State Changes”

*Remember*Your “What Would Life Be Like?”

paper is due next week!