Chapter 19Chemical ThermodynamicsJohn D. BookstaverSt. Charles Community CollegeSt. Peters, MO2006, Prentice Hall, Inc.Modified by S.A. Green, 2006Modified by D. Amuso 2011

ThermodynamicsThe study of the relationships between heat, work, and the energy of a system.

First Law of ThermodynamicsYou will recall from Chapter 5 that energy cannot be created nor destroyed.Therefore, the total energy of the universe is a constant.Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.

Second Law of Thermodynamics

The total Entropy of the universe increases in any spontaneous, irreversible process.

EntropyEntropy can be thought of as a measure of the randomness or disorder of a system.It is related to the various modes of motion in molecules (microstates). Vibrational Rotational Translational

EntropyMolecules have more entropy (disorder) when:Phase Changes from: S L G Example: Sublimation CO2(s) CO2(g) Dissolving occurs (solution forms): Example: NaCl(s) Na+(aq) + Cl-(aq)

Entropy3) Temperature increases Example: Fe(s) at 0 oC Fe(s) at 25 oC 4) For Gases ONLY, when Volume increases or Pressure decreases Examples: 2 Liters He(g) 4 Liters He(g) 3 atm He(g) 1 atm He(g)

Entropy5) Rx results in more molecules/moles of gas Examples: 2NH3(g) N2(g) + 3H2(g) CaCO3(s) CaO(s) + CO2(g)

N2O4(g) 2 NO2(g) This one is difficult to predict: N2(g) + O2(g) 2 NO(g)

EntropyWhen there are more moles Example: 1 mole H2O(g) 2 moles H2O(g)

7) When there are more atoms per molecule Examples: 1 mole Ar(g) 1 mole HCl(g)

1 mole NO2(g) 1 mole N2O4(g)

Entropy8) When an atom has a bigger atomic number 1 mole He(g) 1 mole Ne (g)

Spontaneous ProcessesSpontaneous processes are those that can proceed without any outside intervention.The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously move to just one vessel.

Spontaneous ProcessesProcesses that are spontaneous at one temperature may be nonspontaneous at other temperatures.Above 0C it is spontaneous for ice to melt.Below 0C the reverse process is spontaneous.

Spontaneous ProcessesProcesses that are spontaneous in one direction are nonspontaneous in the reverse direction.Spontaneous processes are irreversible.

Second Law of ThermodynamicsA reversible process results in no overall change in Entropy while an irreversible, spontaneous process results in an overall increase in Entropy.Reversible:Irreversible (Spontaneous):

Third Law of Thermodynamics

The Entropy of a pure crystalline substance at absolute zero is zero. why?

Third Law of Thermodynamics

Standard EntropiesStandard Conditions: 298 K, 1 atm, 1 Molar

The values for Standard Entropies (So) are expressed in J/mol-K. Note: Increase with increasing molar mass.

Standard EntropiesLarger and more complex molecules have greater entropies.

Entropy ChangesDSo = S n Soproducts - S m SoreactantsBe careful: Sunits are in J/mol-K

Note for pure elements:

Gibbs Free Energy Use G to decide if a process is spontaneous

G = negative value = spontaneous G = zero = at equilibrium G = positive value = not spontaneous

Note: equation can be used w/o the o too. DGo = DHo TDSo

Gibbs Free EnergyIf DG is negative DG = maximum amt of energy free to do work by the reaction

2. If G is positive DG = minimum amt of work needed to make the reaction happen DGo = DHo TDSo

Gibbs Free Energy

In our tables, units are: DGo = kJ/mol DHo = kJ/mol DSo = J/mol-K

DGo = DHo TDSo

Free Energy and TemperatureThere are two parts to the free energy equation: H the enthalpy term TS the entropy term

The temperature dependence of free energy comes from the entropy term.

What causes a reaction to be spontaneous?Think Humpty DumptySystem tend to seek: Minimum Enthalpy Exothermic Rx, DH = negative Maximum Entropy More disorder, DS = positiveBecause: DGo = DHo TDSo - = (-) - (+)

Free Energy and TemperatureBy knowing the sign (+ or -) of S and H, we can get the sign of G and determine if a reaction is spontaneous.

At Equilibrium

DGo = zero Therefore: DHo = TDSo

Or: DSo = DHo T

Use this equation when asked to calculate enthalpy of vaporization or enthalpy of fusion.

DGo = DHo TDSo

DGo = S n Gof products - S m Go f reactantsStandard Free Energy Changes

Be careful: Values for Gf are in kJ/mol G can be looked up in tables or calculated from S and H. DGo = DHo TDSo

Free Energy and EquilibriumRemember from above:If DG is 0, the system is at equilibrium.So DG must be related to the equilibrium constant, K (chapter 15). The standard free energy, DG, is directly linked to Keq by:

DGo = - RT ln K Where R = 8.314 J/mol-K

DGo = - RT ln K

If the free energy change is a negative value,the reaction is spontaneous, ln K must be a positive value, and K will be a large number meaning the equilibrium mixture is mainly products.

If the free energy change is zero, ln K = zero and K = one.

Relationships

Free Energy and EquilibriumUnder non-standard conditions, we need to use DG instead of DG.

Q is the reaction quotient from chapter 15.

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