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John W. MooreConrad L. StanitskiPeter C. Jurs

Stephen C. Foster • Mississippi State University

http://academic.cengage.com/chemistry/moore

Chapter 17Thermodynamics: Directionality of

Chemical Reactions

Why are equilibria product- orreactant- favored?

Why do some reactions occurspontaneously?

Why do others require help (heat,spark…)?

Exothermic reactions are oftenproduct favored (but not always).

Reactant- & Product-Favored Processes

2 Al(s) + 3 Br2(ℓ) → 2 AlBr3(ℓ)

Product favored; spontaneous

“Energy will spread out (disperse) unless it is hinderedfrom doing so”.

Dispersal of energyDispersal of energy is probability driven:

• E is more likely to spread over many atoms thanover just a few.

Chemical Reactions & Dispersal of Energy

Most exothermicexothermic reactions are product favored.

• E is transferred to the surroundings.

• Bond (potential) E dispersed into many more atomsand molecules.

E packet

A** B C D A* B* C D A* B C* D

A* B C D* A B** C D A B* C* D

A B* C D* A B C**D A B C* D*

A B C D**

Probability and Dispersal of Energy

A*B* + C DExcited atoms(A, B)

What are the possible outcomes if 2 excited atomsmake contact with 2 more atoms?

E stays with original atoms in only 3 (of 10) cases:

• 70% chance that E will disperse.

• More atoms: E concentration is muchmuch less likely.

Unexcitedatoms

E is dispersed when a system expands.

vacuum

Br2(g)

removablebarrier

• Gases expand to fill a container.

Dispersal of Matter

• Quantum theory: E-levelsfor gas motion get closertogether when volumeexpands.

• More E-levels becomeaccessible at a given T.

• E will disperse over morelevels.

In summaryIn summary:Processes are product favored if E isdispersed from a few levels into manylevels.

Measuring Dispersal of E: Entropy

In thermodynamics, nanoscale E dispersal is measuredby the EntropyEntropy ((S)S) of a system.

This typically occurs during:• exothermic reactions (bond E

dispersal).• processes in which matter is

dispersed.

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At constant T:

ΔS = Sfinal – Sinitial =qrev

T

“rev”: OnlyOnly applies to reversiblereversible changes.

Absolute T

A reversible process:

• can be reversed by a slight change in conditions.

• e.g. ice melting at 0.0°C and 1 atm is reversible.

small decrease in T will cause it to refreeze.

Measuring Dispersal of E: Entropy

Calculate ΔS when 25.0 g of Al(s) melts at its normalmelting point (660.3 °C; 1 atm). ΔHfus(Al)=10.7 kJ mol-1

T = 660.3 °C = 933.5 K

Measuring Dispersal of E: Entropy

°

ΔS =qrev

T=

ΔHfus

Tqp = ΔH

Reversible ifnormal melting

25.0 g

26.98 g mol-1nAl = = 0.9266 mol

ΔS =0.9266 mol(10.7 kJ mol-1)

933.5 K= 0.01062 kJ K-1

ΔS = 10.6 J K-1

A perfect crystal, at absolute zero (0 K) has:

• Minimum molecular motion (min. E dispersal).

SS = 0= 0 (perfect crystal at 0 K)(perfect crystal at 0 K)

(The 3rd law of thermodynamics)

ΔS = Sfinal – Sinitial =qrev

T

Sroom T = ΔS – Sinitial = ΔS

Measure the heat to change from 0 K room-T(reversibly). Gives ΔS and the absoluteabsolute S at room T.

Absolute Entropy Values

Standard molar entropiesStandard molar entropies (S°), are measured byheating 1 mol of substance from 0 K to:

A specified T (often 25°C = 298.15 K)

At constant P = 1 bar

S° has J K-1mol-1 units.

S° is alwaysalways positive, for allall materials.

Absolute Entropy Values

Compound S° (J K-1 mol-1) Compound S° (J K-1 mol-1)

C (graphite) 5.740 Cl2(g) 223.066

C(g) 158.096 Br2(ℓ) 152.231

CH4(g) 186.264 I2(s) 116.135

C2H6(g) 229.60 HCl(g) 186.908

C3H8(g) 269.9 H2O(g) 188.825

C2H4(g) 219.4 H2O(ℓ) 69.91

CH3OH(ℓ) 126.8 NaCl(s) 72.13

CO(g) 197.674 KOH(s) 78.9

CO2(g) 213.74 NaCl(aq) 115.5

F2(g) 202.78 KOH(aq) 91.6

Standard Molar Entropies (298.15 K)

Sgas >> Sliquid > Ssolid

Solid molecules vibrate about fixed lattice points.

Liquid molecules can slide past each other.

Gas molecules have few restrictions on motion.

More motion = higher S

gas liquid solid

S°(I2) 261 - 116

S°(Br2) 245 152 -

S°(Cl2) 223 - -

(J K-1 mol-1)

Qualitative Guidelines for Entropy

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Species type S°(J K-1 mol-1) Species mass S°(J K-1 mol-1)

CH4 alkane 186 Ar 40 155

C2H6 alkane 230 CO2 44 214

C3H8 alkane 270 C3H8 44 270

Larger, more complex, molecules have larger S:

S°complex molecule > S°simple molecule

Larger molecules have more ways to distribute E(more bonds to hold potential and kinetic E).

Qualitative Guidelines for Entropy

S°(J K-1mol-1) Comment

MgO 27 +2 and -2 charges

NaF 51 +1 and -1 charges

NaCl 72 Cl- much larger than F-, less attraction.

Ionic solidsIonic solids

Weaker ionic forces = larger S

(Less tightly bound = easier motion)

Qualitative Guidelines for Entropy

Solid or liquid dissolvingSolid or liquid dissolving

Larger matter dispersal. S usually* increases:

*Strong solvation may increase order. ΔS < 0 ispossible.

Qualitative Guidelines for Entropy

Substance S° (pure) S° (aq)

CH3COOH(ℓ) 160 179

NH4NO3(s) 151 260

J K-1 mol-1 units

Gas dissolvingGas dissolving

Gas-molecule motion becomes restricted. ΔS < 0.

Qualitative Guidelines for Entropy

Substance S° (gas) S° (aq. soln)

CO2 214 118

H2 131 58

CH3OH 240 133

J K-1 mol-1 units

Predicting Entropy Changes

CaCO3(s) → CaO(s) + CO2(g) ΔS = 161 J K-1mol-1

Large entropy increase:

• Solid → gas, ΔS > 0.

• 1 molecule → 2 molecules, ΔS > 0.

H2(g) + F2(g) → 2 HF(g) ΔS = 14 J K-1mol-1

• All gases, S ≈ constant.

• 2 reactant gases, 2 product gases, S ≈ constant.

Actual ΔS shows a small increase.

Predict whether S will increase or decrease for:

2 CO(g) + O2(g) → 2 CO2(g)

DecreaseDecrease 3 gas molecules → 2 gas molecules.

NaCl(s) → Na+(aq) + Cl-(aq)

IncreaseIncrease Ions locked in a crystal → ions dispersed in water.

H2O(ℓ) → H2O(s)

DecreaseDecrease Liquid molecules less restricted than solidmolecules.

Predicting Entropy Changes

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Entropy is a state function (like H). For a reaction:

ΔS° = (nproductS°product) – S(nreactantS°reactant)

ExampleExample

Methanol is a common fuel additive. Calculate ΔS°

at 298 K for its combustion:2 CH3OH(ℓ) + 3 O2(g) → 2 CO2(g) + 4 H2O(ℓ)

Look up S° values (J K-1 mol-1):

CH3OH(ℓ) = 126.8 O2(g) = 205.138

CO2(g) = 213.74 H2O(ℓ) = 69.91

Calculating Entropy Changes Calculating Entropy ChangesExample:: ΔS° at 298 K for: 2 CH3OH(ℓ) + 3 O2(g) → 2 CO2(g) + 4 H2O(ℓ)

ΔS° = (nproductS°product) – (nreactantS°reactant)

= (2S°CO2 + 4S°H2O) – (2S°CH3OH + 3S°O2)

= [2(213.74) + 4(69.91)] – [2(126.8) + 3(205.14)]

= 707.12 – 869.02

= –161.90 J K-1 mol-1

S decreasesdecreases as expected: 3 gas + 2 ℓ → 2 gas + 4 ℓ

Entropy & the 2nd Law of Thermodynamics

E dispersal is accompanied by an increase in disorderof a system:

Increased disorder = increased S

Second Law of ThermodynamicsSecond Law of Thermodynamics

“The total entropy of the universe is continuouslyincreasing.”

• The universe is slowly becoming more disordered.

The entropy of a system can increase oror decrease.

• Suniverse always goes up.

• Suniverse = Ssystem + Ssurr (No subscript? S = Ssystem ).

• If Ssystem falls, Ssurr must increase by a larger amount.

Entropy & the 2nd Law of Thermodynamics

ΔS°surr = = =qrev

T

ΔHsurr

T

-ΔHsys

T

Entropy & the 2nd Law of Thermodynamics

Calculate ΔSuniv for the combustion of octane in air:

2 C8H18(g) + 25 O2(g) 16 CO2(g) + 18 H2O(ℓ)

For the reaction ΔH° = -11,024 kJ and ΔS° = -1383.9

J/K at 298.15 K.

°

ΔSuniv = ΔSsys + ΔSsurr

ΔSuniv = -1383.9 J/K + (++11,024 x 103 J)/298.15 K

= -1383.9 J/K + 36.975 kJ/K

= +35,591 J/K

° ° °

given

-ΔHsys / T

Gibbs Free EnergyNeither entropy (S), nor enthalpy (H ), alone can predictwhether a reaction is product favored.

Spontaneous (product favored) reactions can:

• Be exothermic or endothermic.

• Increase or decrease the Ssystem.

The Gibbs free energy (Gibbs free energy (GG)), combines H and S.

ΔG doesdoes predict if a reaction is product favored or not.

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Gibbs Free Energy

For a constant T change, ΔG is equal to:

ΔΔGG == ΔΔHH -- TTΔΔSS

If G:

DecreasesDecreases (ΔG < 0) a reaction is product favoredproduct favored

IncreasesIncreases (ΔG > 0) a reaction is reactant favoredreactant favored

(at constant P and T).

ΔG° can be calculated from ΔH° and ΔS° values…

… or from Gibbs free energies of formationGibbs free energies of formation (ΔGf°).

ΔG° = (nproductΔGf°product) – (nreactantΔGf°reactant)

ΔGf° equals:

• ΔGreaction to make 1 mol of compound from itselements.

• 0 for an element in its most stable state (like ΔHf° )

Gibbs Free Energy

Calculate ΔG° for the following reaction at 25°C:

CO(g) + 2 H2(g) → CH3OH(ℓ)

ΔG° = ΔGf°CH3OH – (2ΔGf°H2 + 1ΔGf°CO)

= -166.3 kJ/mol – [2(0) + 1(-137.2 kJ/mol)]

= -29.1 kJ/mol

Product favoredProduct favored

ΔGf° values: CO(g) = -137.2 kJ/mol

CH3OH(ℓ) = -166.3 kJ/mol

Gibbs Free Energy

element in itsstandard state

Reactions go from reactant to product-favored when:ΔG = 0 = ΔH − TΔS

or when: ΔH = TΔS

or T = ΔHΔS

Effect of T on Reaction Direction

Sign of ΔH Sign of ΔS Product Favored?

Negative (exothermic) Positive Yes

Negative (exothermic) Negative Yes at low T; No at high T

Positive (endothermic) Positive No at low T; Yes at high T

Positive (endothermic) Negative No

Since ΔG = ΔH – TΔS

Effect of T on Reaction DirectionAt what T will the following reaction be productfavored? CH4(g) + H2O(g) CO(g) + 3 H2(g)

ΔHf° (kJ/mol) −74.81 −241.82 −110.53 0

S° (J K-1 mol-1) 186.26 188.83 197.67 130.68

ΔS°=[(197.67) +3(130.68)] − [186.26 + 188.83]J K-1mol-1

= 214.62 J K-1 mol-1

ΔH° = [(-110.53) + 3(0)] − [(-74.81) + (-241.82)] kJ mol-1

= 206.10 kJ mol-1

It will switch direction at:

T = = = 960.30 K2.0610 x 105 J mol-1

214.62 J K-1 mol-1ΔHΔS

Effect of T on Reaction Direction

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Gibbs free energy G and the equilibrium constant Kboth show if a reaction is product-favored and arerelated.

ΔΔGG°° = −= −RTRT lnln KK°°

K° must be unitless.

Each quantity in the K° expression is divided by itsstandard state value:

Gibbs Free Energy and Equilibrium Constants

• concentrations by 1 mol L-1 (solids and liquids).

• pressures by 1 bar (gas reactions).

ΔG° = ΔGf°(N2O4) - 2ΔGf°(NO2)

= 97.89 – 2(51.31) kJ/mol = -4.73 kJ/mol

Evaluate K° (298 K) for the reaction:

2 NO2(g) N2O4(g)

ΔG°RT

-4.73 x 103 J/mol(8.314 J K-1 mol-1)(298 K)

ln K° = – = – = 1.91

K° = 6.75

Gibbs Free Energy and Equilibrium Constants

Gibbs Free Energy & Maximum Work

ΔG = maximum usefuluseful work that can be done by a

reaction on its surroundings (constant T and P).

ΔG = wsystem = - wmax

“free” energy = available energy

For:

2 H2(g) + O2(g) 2 H2O(g) ΔG°= -474.3 kJ

Every 2 moles of H2 consumed can do up to 474.3 kJof work.

Gibbs Free Energy & Maximum Work

ΔG also equals the minimumminimum work required to cause a

reactant-favored process to occur.

2 H2O(g) 2 H2(g) + O2(g) ΔG°= 474.3 kJ

A minimumminimum of 474.3 kJ of work must be used to

produce 2 mol of H2 from liquid water.

Thermodynamically stableThermodynamically stable

Reaction is not product-favored.

• Thermodynamically stable:

Pd + O2 → PdO2 ΔG° = +326 kJ/mol

• Thermodynamically unstable:

4 Al + 3 O2 → 2 Al2O3 ΔG° = -3164.6 kJ/mol

Kinetically stableKinetically stable

Product-favored, but too slow to be important.

• Diamond is thermodynamically unstable

Cdiamond → Cgraphite ΔG° = -2.9 kJ/mol

• Kinetically stable, too slow to be important.

• Ea is too large

Thermodynamic and Kinetic Stability