Patrick Riley , Robert Hanson, Paul Fischer and Jeff Schwinefus

transcript

Demos for Free (Energy, that is..)Chemical Demonstrations Explained

with Graphs of Free Energy vs. Temperature

Patrick Riley, Robert Hanson, Paul Fischer and Jeff Schwinefus

Department of Chemistry, St. Olaf College

Department of Chemistry, Macalaster College

July 19, 2004

“Demo Day”

A skit is performed to demonstrate recrystallization. Can we recover our “contaminated” goods?

Overall Concept Map For Molecular Thermodynamics

internal energy, U

heat, q

temperature, Tentropy, S enthalpy, H

work, w

free energy, G

equilibrium constant, Kreaction quotient, Q

Our Focus Today: Free Energy and Temperature

internal energy, U

heat, q

temperature, Tentropy, S enthalpy, H

work, w

free energy, G

equilibrium constant, Kreaction quotient, Q

Free Energy vs. Reaction Coordinate Graphs

Reactants

Products

Equilibrium mixture

Course of reaction

Equilibrium is the state where free energy is minimized.

Free Energy vs. Temperature Graphs

G = H – TS

reactants

products

Equilibrium is the state where G = 0.

G = H – TS

reactants

products

ΔH < 0

G = H – TS

reactants

products

ΔH < 0

slope = -S

Sproducts < Sreactants

ΔS < 0

Applications: Use of Classroom Demonstrations

Demonstration 1: “Balloon in an Erlenmeyer”

H2O (l) H2O (g)

Qualitative illustration of the interdependence between vapor pressure and temperature.

Questions discussed during class include, “What is inside the flask?” before and after boiling.

G = H – TS

H2O(g)

H2O(l)

P = 1 atm

G = H – TS

P2 > 1 atm

H2O(g)

H2O(l)

G = H – TS

H2O(g)

H2O(l)

P = 1 atm

G = H – TS

P << 1 atm

Troom T1

H2O(g)

H2O(l)

Demonstration 2: “Can of Beans”

H2O (l) H2O (g)

G = H – TS

H2O (l)

PH2O < 1 atm

H2O (g)Will the can of beans survivethe extreme heat of the hotplate?Get ready to duck!

H2O (l) H2O (g)

G = H – TS

H2O (l) PH2O = 1 atm

H2O (l) H2O (g)

G = H – TS

H2O (l)PX > 1 atm

Note: stirring, not heat turned on for the purpose of thisdemonstration -- works every time!

Demonstration 3: Low Pressure

Evaporation CH2Cl2 (l) CH2Cl2 (g)

G = H – TS

P << 1 atm

Teq Troom

Boiling observed with no application of heat.

Demonstration 3: Low Pressure

Evaporation CH2Cl2 (l) CH2Cl2 (g)

G = H – TS

P << 1 atm

Teq T2

Boiling observed with no application of heat. Frost formation on flask discussed as result of the endothermic (liquid gas) reaction.

Demonstration 4: Relative Humidity and Dew Point

Current weather data from Faribault airport is taped and played for the class.

G = H – TS

H2O (g)

H2O (l)

Pvap = P100%

Relative humidity relates the ratio of the actual water vapor pressure in the air to the equilibrium vapor pressure for the current air temperature.

G = H – TS

H2O (g)

H2O (l)

Tcurrent

Pvap = 0.30 x P100%

Relative humidity <100% indicates liquid water is not generally at equilibrium with atmospheric water vapor; vaporization is thermodynamically favored.

G = H – TS

H2O (g)

H2O (l)

Tcurrent

Pvap = 0.30 x P100%

The dew point (Tdp) indicates where the actual pressure water vapor curve crosses the liquid curve.

Demonstration 5: Triple Point of CO2

G = H – TS

–78 oC

CO2 (l)CO2 (s)

CO2 (g)

P = 1 atm

Liquid phase is always higher in free energy than solid or gas and therefore is not observed.

Demonstration 5: Triple Point of CO2

G = H – TS

–56 oC

CO2 (l)CO2 (s)

CO2 (g)

P = 5.11 atm

Liquid phase is always higher in free energy than solid or gas and therefore is not observed. Increasing the pressure results in a flatter gas curve which crosses the solid and liquid curves at the melting point.

Demonstration 6: Solubility or “Hot Lemonade Tastes Better”

Sugar (s) Sugar (aq)

Endothermic dissolution of solids emphasized.

Endothermic dissolution of solids emphasized. Curves cross when saturated.

G = H – TS

sugar (aq)

sugar (s)

[sugar] low

Endothermic dissolution of solids emphasized. Curves cross when saturated. Increase in temperature results in dissolution.

G = H – TS

sugar (aq)

sugar (s) [sugar] high

Demonstration 7: Supersaturated Sodium Acetate Solutions

NaOAc (s) NaOAc (aq) The molar free energy of the aqueous solute is higher than that of the solid in a supersaturated solution. Precipitation is thermodynamically favored to equalize the molar free energy of the species; external stimulus required.

G = H – TS

NaOAc (aq)

NaOAc (s)

TsatTact

Demonstration 8: Dissolution of Gases or “Dead Fish”

O2 (g) O2 (aq)

G = H – TS

O2 (g)

O2 (aq)

T > TeqTeq

Increase in temperature favors solute gas; solution is degassed.

Demonstration 9: Freezing Point Depression or “Fool the Waitperson”

As the solute dissolves liquid water becomes impure and its entropy increases.

G = H – TS

H2O (l)H2O (s)

H2O (g)

G = H – TS

Tmp < 0 oC

H2O (l)H2O (s)

H2O (g)

The liquid water curve’s slope becomes more steep and establishes a lower crossing temperature.

G = H – TS

Tmp lower

H2O (l)H2O (s)

H2O (g)

Tbp higher

Note that boiling point elevation arises from the same phenomenon. Increasing molar entropy of the liquid will push the boiling point to the right.

Laboratory Exercise: “Chemical Magic: Free Energy”

Why should chemistry professors have all the fun?

Students perform a demonstration in groups of 2-3,

answer 4-6 questions about the demonstration then

move on to the next station.

G-T graphs are drawn for each laboratory exercise.

Laboratory demonstrations include:“Inside-Out Balloon”

“Fog Chamber”“Cold Boiling”“Liquid Air”“Cool Glue”“Hot Ice”“Crystals from not-so-thin-air”“Instant Snowflakes”“Instant Hot; Instant Cold”“Magic Rope”

Upon completing the third demonstration (45 minutes) the studentsprepare to present a magic show for their classmates.

“Chemical Magic is fun!”

Conclusion

Free energy vs. temperature graphs can be utilized to discuss a variety of common classroom demonstrations.

“Chemical Magic” laboratory exercise can effectively supplement the discussion initiated during “Demo-Days.”

Patrick Riley , Robert Hanson, Paul Fischer and Jeff Schwinefus

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