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Chapter 19Chemical Thermodynamics

Entropy and free energyLearning goals and key skills: Understand the meaning of spontaneous process, reversible process, irreversible process,

and isothermal process. State the second law of thermodynamics. Describe the kinds of molecular motion that a molecule can possess. Explain how the entropy of a system is related to the number of accessible microstates. Predict the sign of S for physical and chemical processes. State the third law of thermodynamics. Calculate standard entropy changes for a system from standard molar entropies. Calculate entropy changes in the surroundings for isothermal processes. Calculate the Gibbs free energy from the enthalpy change and entropy change at a given

temperature. Use free energy changes to predict whether reactions are spontaneous. Calculate standard free energy changes using standard free energies of formation. Predict the effect of temperature on spontaneity given H and S. Calculate G under nonstandard conditions. Relate G°and equilibrium constant.

Energetics of solutionsAn endothermic process is not favored based on the change in enthalpy, yet they occur. Why?

We must also look at the disorder (entropy, S) of the system.

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0th Law of ThermodynamicsIf A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then C is also in thermal equilibrium with A.

1st Law of ThermodynamicsEnergy of the universe is constant (conserved). U = q + wq = heat absorbed by the system, w = work done on the system

2nd Law of ThermodynamicsIn a spontaneous process, the entropy of the universe increases.

∆Suniverse = ∆Ssys + ∆Ssurr > 0 (if spontaneous)

3rd Law of ThermodynamicsThe entropy of a pure crystalline substance at absolute zero is zero: S(0 K) = 0.

Review Chapter 5: energy, enthalpy, 1st law of thermo

Thermodynamics: the science of heat and work

Thermochemistry: the relationship between chemical reactions and energy changes

Energy (E) The capacity to do work or to transfer heat.

Work (w) The energy expended to move an object against an opposing force.

w = F d

Heat (q) Derived from the movements of atoms and molecules (including vibrations and rotations).

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-q, heat evolved by thesystem (exothermic)

+q, heat absorbed by system (endothermic)

System∆E = q + w

+w, work done onto the system

-w, work done by the system

Surro

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1st law of thermodynamics: the law of conservation of energy

Energy is neithercreated nor destroyed

Internal Energy = heat + workDE = q + w

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Enthalpy, H• the heat content of a substance at constant pressure.

• a state function

• an extensive property

• reversible

The enthalpy change, H, is defined as the heat gained or lost by the system under constant pressure; it depends upon the identity and states of the reactants and products.

DH = qp

Standard Enthalpy ValuesMost ∆H values are labeled ∆Ho

Measured under standard conditionsP = 1 atmT = usually 298.15 K (25 oC)(Concentration = 1 M)

∆Hfo = standard molar enthalpy of formation

(Appendix C)

∆Horxn = ∆Hf

o (products) - ∆Hf

o (reactants)∆Ho

rxn = ∆Hfo (products)

- ∆Hfo (reactants)

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Spontaneous Reactions: reactions that occur without outside intervention

In general, spontaneous reactions are exothermic.

Thermite reaction:Fe2O3(s) + 2 Al(s) →

2 Fe(s) + Al2O3(s)∆Hrxn = -848 kJ

But many spontaneous reactions or processes are endothermic or even have ∆H = 0.

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Spontaneous Processes

• Spontaneous 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 return to vessel B.

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Which has a positive value of S?

DS = Sfinal Sinitial

Spontaneous Processes

Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

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Spontaneous Processes• Processes 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.

Reversible Processes

In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.

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Irreversible Processes

• Irreversible processes cannot be undone by exactly reversing the change to the system.

• Spontaneous processes are irreversible.

EntropyFor a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the absolute temperature:

Like energy and enthalpy, entropy is a state function and

DS = Sfinal Sinitial

Ssys =qrevT

For any spontaneous process, the entropy of the universe increases: (Suniv >0)

Ssurr = =-DHsys

T-qsys

T(at constant P,T)

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2nd Law of Thermodynamics

For reversible processes:

Suniv = Ssystem + Ssurroundings = 0

For irreversible processes:

Suniv = Ssystem + Ssurroundings > 0

Entropy on the molecular scale

Molecules exhibit several types of motion:– Translational: Movement of the entire molecule from one

place to another.– Vibrational: Periodic motion of atoms within a molecule.– Rotational: Rotation of the molecule on about an axis or

rotation about bonds.

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open the stopcock

no work (w = 0)no heat (q = 0)

spontaneous, nonetheless.

Reverse is unimaginable.

Statistical treatment of entropy

# particles probability they are all in A2 1/4 = 0.253 1/8 = 0.1254 1/16 = 0.062510 1/1024 = 0.0009765625

N = 6.022 × 1023 2-N ≈ 0

Entropy: matter dispersal

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A key contribution is the dispersal of energy over many different energy states (each state is called a microstate).

(This is similar to matter dispersal.)

Let W = number of microstates of a system

S = k ln W

where k = 1.38 × 10-23 J/K (Boltzmann constant)

Entropy: energy dispersal

Entropy on the Molecular Scale

• The change in entropy for a process:

S = k lnWfinal k lnWinitial

WfinalWinitial

S = k ln

• Entropy increases with the number of microstates in the system.

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S = k ln WIn general, the number of microstates available to a system increases with

an increase in volume, an increase in temperature, or an increase in the number of molecules

because any of these changes increases the possible positions and energies of the molecules of the system.

Entropy and microstates

Solutions

Generally, when a solid is dissolved in a solvent, entropy increases.

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In general,• S (solids) < S (liquids) << S (gases)

• S (small molecules) < S (large molecules)• S (simple molecules) < S (complex molecules)• S increases as the temperature is raised

So (J/K•mol)

H2O (liq) 69.91

H2O (gas) 188.8

So (J/K•mol)

H2O (liq) 69.91

H2O (gas) 188.8

Entropy, S

Entropy Changes

• In general, entropy increases when– Gases are formed from

liquids and solids;– Liquids or solutions

are formed from solids;

– The number of gas molecules increases;

– The number of moles increases.

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Third Law of ThermodynamicsThe entropy of a pure crystalline substance at absolute zero is zero: S(0 K) = 0.

S = k ln W

Standard Entropies

• These are molar entropy values of substances in their standard states.

• Standard entropies tend to increase with increasing molar mass.

See Appendix C.

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Standard Entropies

Larger and more complex molecules have greater entropies.

• Boltzmann Equation: S = k ln Wk = 1.38 × 10-23 J/K

• ΔS = qrev / T

• ΔS°sys = Σ nS°

(products) – Σ mS°(reactants)

• ΔS°univ = Σ S°

sys + Σ S°surr

• ∆Suniverse = ∆Ssys + ∆Ssurr > 0(for a spontaneous reaction)

Entropy equations

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Entropy Change in the Universe

Since Ssurroundings = and qsystem = Hsystem

This becomes:Suniverse = Ssystem +

Multiplying both sides by T, we getTSuniverse = Hsystem TSsystem

HsystemT

qsystemT

∆Suniverse = ∆Ssystem + ∆Ssurroundings

Gibbs Free Energy

• TSuniverse is defined as the Gibbs free energy, G.

• When Suniverse is positive, G is negative.

• Therefore, when G is negative, a process is spontaneous.

DG = DH TDS

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Gibbs Free Energy

1. If G is negative, the forward reaction is spontaneous.

2. If G is 0, the system is at equilibrium.

3. If G is positive, the reaction is spontaneous in the reverse direction.

Standard Free Energy Changes

Analogous to standard enthalpies of formation are standard free energies of formation, G.f

G = SnG (products) SmG (reactants)f f

where n and m are the stoichiometric coefficients.

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Free Energy Changes

At temperatures other than 25°C,G = H TS

How does G change with temperature?

• There are two parts to the free energy equation:H the enthalpy termTS the entropy term

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

Free Energy and TemperatureGibbs free energy change =

total energy change for system - energy lost in disordering the system

If ∆Go is negative, the reaction is spontaneous (andproduct-favored).

If ∆Go is positive, the reaction is not spontaneous (andreactant-favored).

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Calculating ΔG°

∆Gorxn = ∆Gf

o (products) - ∆Gfo (reactants)∆Go

rxn = ∆Gfo (products) - ∆Gf

o (reactants)

a) Determine ∆Ho and ∆So and use Gibbsequation. ∆Go = ∆Ho - T∆So

b) Use tabulated values of free energies offormation, ∆Gf

o.

Determine G°rxn for the following combustion reaction.CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

Hf° (kJ/mol) -74.87 0 -393.509 -285.83S° (J/mol K) 186.26 205.07 213.74 69.95Gf° (kJ/mol) -50.8 0 -394.359 -237.15

H°rxn = -890.30 kJ exothermicS°rxn = -242.76 J/K more orderedG°rxn = ∆Ho

rxn - T∆Sorxn = -817.92 kJ spontaneous

G°rxn = -817.92 kJ spontaneous

Example: Gibbs free energy

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A + B → C + D irreversible (easy Gen Chem I)

A + B C + D reversible (most reactions)

forward reaction = reverse reaction

No drive to make all products (spontaneous reaction) or all reactants (nonspontaneous)

Examples: melting point boiling pointA(s) A (l) A(l) A (g)

When G = 0, equilibrium

Given ice or H2O (s) Hf° = -292.70 kJ/molS° = 44.8 J/mol·K

water or H2O (l) Hf° = -285.83 kJ/molS° = 69.95 J/mol·K

Find the melting point of water.

It better be 0 °C!

Example: melting point of water

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Free Energy and Equilibrium

Under any conditions, standard or nonstandard, the free energy change can be found this way:

G = G + RT lnQ

(Under standard conditions, all concentrations are 1 M, so Q = 1 and lnQ = 0; the last term drops out.)

Free Energy and Equilibrium

• At equilibrium, Q = K, and G = 0.• The equation becomes

0 = G + RT lnK• Rearranging, this becomes

G = RT lnKor,

K = e-GRT

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Use tabulated* free energies of formation to calculated the equilibrium constant for the following reaction at 298 K.

N2O4(g) 2NO2(g)

Example: Gibbs free energy and equilibrium constant

*See Appendix C, textbook pages 1059-1061