Chapter 16Chapter 16• Thermodynamics: Entropy,Free Energy,and Equilibrium

• Thermodynamics: Entropy,Free Energy,and Equilibrium

spontaneous

nonspontaneous

In this chapter we will determine the direction of a chemical reaction and calculate equilibrium constant using thermodynamic values: Entropy and Enthalpy.

Spontaneous ProcessesSpontaneous Processes

Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence.

Universe: System + SurroundingsUniverse: System + Surroundings

The system is what you observe; surroundings are everything else.

Thermodynamics

State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

energy, pressure, volume, temperature

Laws of Thermodynamics Laws of Thermodynamics

1) Energy is neither created nor destroyed.

2) In any spontaneous process the total Entropy of a system and its surrounding always increases. We will prove later: ∆G = ∆Hsys – T∆Ssys

∆G < 0 Reaction is spontaneous in forward direction3) The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0)

Enthalpy, Entropy, and Spontaneous ProcessesEnthalpy, Entropy, and Spontaneous Processes

State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state.

Enthalpy Change (H): The heat change in a reaction or process at constant pressure.

Entropy (S): The amount of Molecular randomness change in a reaction or process at constant pressure.

Enthalpy ChangeEnthalpy Change

= +40.7 kJHvap

CO2(g) + 2H2O(l)CH4(g) + 2O2(g)

H2O(l)H2O(s)

H2O(g)H2O(l)

2NO2(g)N2O4(g)

= +3.88 kJH°

= +6.01 kJHfusion

= -890.3 kJH°

= +57.1 kJH°

Endothermic:

Exothermic:

Na1+(aq) + Cl1-(aq)NaCl(s)H2O

Entropy ChangeEntropy Change

S = Sfinal - Sinitial

EntropyEntropy

EntropyEntropy

The sign of entropy change, S, associated with the boiling of water is _______.

1. Positive

2. Negative

3. Zero

Correct Answer:

Vaporization of a liquid to a gas leads to a large increase in volume and hence entropy; S must be positive.

1. Positive

2. Negative

3. Zero

Entropy and Temperature 02Entropy and Temperature 02

Third Law of Thermodynamics

The entropy of a perfect crystalline substance is zero at the absolute zero ( K=0)

Ssolid < Sliquid < Sgas

Entropy Changes in the System (Ssys)

When gases are produced (or consumed)

• If a reaction produces more gas molecules than it consumes, S0 > 0.

• If the total number of gas molecules diminishes, S0 < 0.

• If there is no net change in the total number of gas molecules, then S0 may be positive or negative .

What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) 2ZnO (s)

The total number of gas molecules goes down, S is negative.

Standard Molar Entropies and Standard Entropies of ReactionStandard Molar Entropies and Standard Entropies of Reaction

Standard Molar Entropy (S°): The entropy of 1 mole of a pure substance at 1 atm pressure and a specified temperature.

Calculated by using S = k ln W , W = Accessible microstates

of translational, vibrational and rotational motions

Review of Ch 8: Review of Ch 8:

• Calculating H° for a reaction:

H° = H°f (Products) – H°f (Reactants)

• For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient.

aA + bB cC + dD

H° = [cH°f (C) + dH°f (D)] – [aH°f (A) + bH°f (B)]

See Appendix B, end of your book

•∆G = ∆H – T∆S see page 301 of your book

We will show the proof of this formula later in this chapter

Entropy Changes in the System (Ssys)

aA + bB cC + dD

S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

S0rxn nS0

(products)= mS0 (reactants)-

The standard entropy change of reaction (S0 ) is the entropy change for a reaction carried out at 1 atm and 250C.

rxn

What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g)

S0(CO) = 197.6 J/K•molS0(O2) = 205.0 J/K•mol

S0(CO2) = 213.6 J/K•mol

S0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]

S0rxn = 427.2 – [395.2 + 205.0] = -173.0J/K•mol

Consider 2 HConsider 2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq)O(liq)

∆∆SSoo = 2 S = 2 Soo (H (H22O) - [2 SO) - [2 Soo (H (H22) + S) + Soo (O (O22)])]

∆∆SSoo = 2 mol (69.9 J/K•mol) - = 2 mol (69.9 J/K•mol) - [2 mol (130.6 J/K•mol) + [2 mol (130.6 J/K•mol) +

1 mol (205.0 J/K•mol)]1 mol (205.0 J/K•mol)]∆∆SSoo = -326.4 J/K = -326.4 J/KNote that there is a Note that there is a decrease in Entropy decrease in Entropy because 3 because 3

mol of gas give 2 mol of liquid.mol of gas give 2 mol of liquid.

Calculating ∆S for a ReactionCalculating ∆S for a ReactionCalculating ∆S for a ReactionCalculating ∆S for a Reaction

∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)

Calculate S° for the equation below using the standard entropy data given:

2 NO(g) + O2(g) 2 NO2(g)

S° values (J/mol-K): NO2(g) = 240, NO(g) = 211, O2(g) = 205.

1. +176 J/mol

2. +147 J/mol

3. 147 J/mol

4. 76 J/mol

Correct Answer:

S° = 2(240) [2(211) + 205]S° = 480 [627]

S° = 147

reactantsproducts SSS

1. +176 J/K

2. +147 J/K

3. 147 J/K

4. 76 J/K

Thermite ReactionThermite Reaction

Spontaneous Processesand Entropy 07Spontaneous Processesand Entropy 07

• Consider the gas phase reaction of A2 molecules (red) with B atoms (blue).

(a) Write a balanced equation for the reaction.(b) Predict the sign of ∆S for the reaction.

Entropy Changes in the Surroundings (Ssurr)

Exothermic ProcessSsurr > 0

Endothermic ProcessSsurr < 0

In any spontaneous process the total Entropy of a system and its surrounding always increases.

Second Law Of Thermodynamic:

Entropy and the Second Law of ThermodynamicsEntropy and the Second Law of Thermodynamics

ssurr - H)

ssurr T

1

ssurr = T

- H

See example of tossing a rock into a calm waters vs. rough watersPage 661 of your book

)

2nd Law of Thermodynamics 012nd Law of Thermodynamics 01

• The total entropy increases in a spontaneous

process and remains unchanged in an equilibrium

process.

Spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0

Equilibrium: ∆Stotal = ∆Ssys + ∆Ssur = 0

• The system is what you observe; surroundings are

everything else.

2 H2 H22(g) + O(g) + O22(g) ---> 2 H(g) ---> 2 H22O(liq) O(liq) ∆H ∆H °° = -571.7 kJ = -571.7 kJ

Sosurroundings =

- (-571.7 kJ)(1000 J/kJ)

298.15 KSo

surroundings = - (-571.7 kJ)(1000 J/kJ)

298.15 K

2nd Law of 2nd Law of ThermodynamicsThermodynamics

∆∆SSoosurroundingssurroundings = +1917 J/K = +1917 J/K

Ssurr Hsys

T

Calculate change of entropy of surrounding in the following reaction:

Spontaneous Reactions 01Spontaneous Reactions 01

• The 2nd law tells us a process will be spontaneous

if ∆Stotal > 0 which requires a knowledge of ∆Ssurr.

spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0

S sur Hsys

T

•∆Stotal = ∆Ssys + ( ) > 0 H sys

T

-T (∆Stotal = ∆Ssys + ( ) ) < 0 H sys

T

Gibbs Free Energy02Gibbs Free Energy02• The expression –T∆Stotal is equated as Gibbs free

energy change (∆G)*, or simply free energy change:

•

•

• ∆G = ∆Hsys – T∆Ssys

• ∆G < 0 Reaction is spontaneous in forward direction.∆G = 0 Reaction is at equilibrium.∆G > 0 Reaction is spontaneous in reverse direction.

-T . ∆Stotal = -T. ∆Ssys + ΔHsys < 0

–T∆Stotal = ΔG

* Driving force of a reaction, Maximum work you can get from a system.

Calculate ∆GCalculate ∆Go o rxn for the following:rxn for the following:Calculate ∆GCalculate ∆Go o rxn for the following:rxn for the following:

CC22HH22(g) + 5/2 O(g) + 5/2 O22(g) --> 2 CO(g) --> 2 CO22(g) + H(g) + H22O(g)O(g)

Use enthalpies of formation to calculateUse enthalpies of formation to calculate ∆∆HHoo

rxnrxn = -1256 kJ = -1256 kJ

Use standard molar entropies to calculate ∆SUse standard molar entropies to calculate ∆Soorxnrxn ( see page ( see page

658, appendix A-10)658, appendix A-10) ∆∆SSoo

rxnrxn = -97.4 J/K or -0.0974 kJ/K = -97.4 J/K or -0.0974 kJ/K

∆ ∆GGoorxnrxn = -1256 kJ - (298 K)(-0.0974 kJ/K) = -1256 kJ - (298 K)(-0.0974 kJ/K)

= -1227 kJ= -1227 kJ

Reaction is Reaction is product-favoredproduct-favored in spite of negative ∆S in spite of negative ∆Soorxnrxn. .

Reaction is Reaction is “enthalpy driven”“enthalpy driven”

1. +33 kJ

2. +66 kJ

3. 66 kJ

4. 33 kJ

Correct Answer:

H° = 92 kJS° = 198.3J/mol . K

G° = H° TS°G° = (92) (298)(0.1983)

G° = (92) + (59.2) = 32.9 KJ

1. +33 kJ

2. +66 kJ

3. 66 kJ

4. 33 kJ

Gibbs Free Energy03Gibbs Free Energy03

Using ∆G = ∆H – T∆S, we can predict the sign of ∆G from the sign of ∆H and ∆S.

1) If both ∆H and ∆S are positive,

∆G will be negative only when the temperature value is large.

Therefore, the reaction is spontaneous only at high temperature.

2) If ∆H is positive and ∆S is negative,

∆G will always be positive.

Therefore, the reaction is not spontaneous

3) If ∆H is negative and ∆S is positive,

∆G will always be negative.

Therefore, the reaction is spontaneous

4) If both ∆H and ∆S are negative,

∆G will be negative only when the temperature value is small.

Therefore, the reaction is spontaneous only at Low temperatures.

Gibbs Free Energy: 04Gibbs Free Energy: 04∆G = ∆H – T∆S

Gibbs Free Energy04Gibbs Free Energy04

Gibbs Free Energy06Gibbs Free Energy06

• What are the signs (+, –, or 0) of ∆H, ∆S, and ∆G for the following spontaneous reaction of A atoms (red) and B atoms (blue)?

Gibbs Free Energy02Gibbs Free Energy02• The expression –T∆Stotal is equated as Gibbs free

energy change (∆G)*, or simply free energy change:

•

•

• ∆G = ∆Hsys – T∆Ssys

• ∆G < 0 Reaction is spontaneous in forward direction.∆G = 0 Reaction is at equilibrium.∆G > 0 Reaction is spontaneous in reverse direction.

-T . ∆Stotal = -T. ∆Ssys + ΔHsys < 0

–T∆Stotal = ΔG

* Driving force of a reaction, Maximum work you can get from a system.

Standard Free-Energy Changes for ReactionsStandard Free-Energy Changes for Reactions

Calculate the standard free-energy change at 25 °C for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes:

S° = -198.7 J/K

2NH3(g)N2(g) + 3H2(g)

= -33.0 kJ

H° = -92.2 kJ

G° = H° - TS°

1000 J

1 kJ

K

-198.7 Jx x=

-92.2 kJ - 298 K

Gibbs Free Energy05Gibbs Free Energy05

• Iron metal can be produced by reducing iron(III)

oxide with hydrogen:

Fe2O3(s) + 3 H2(g) 2 Fe(s) + 3 H2O(g)

∆H° = +98.8 kJ; ∆S° = +141.5 J/K

1. Is this reaction spontaneous at 25°C?

2. At what temperature will the reaction become

spontaneous?

Decomposition of CaCO3 has a H° = 178.3 kJ/mol and S° = 159 J/mol K. At what temperature does this become spontaneous?

1. 121°C2. 395°C3. 848°C4. 1121°C

Correct Answer:

1. 121°C2. 395°C3. 848°C4. 1121°C

T = H°/S°

T = 178.3 kJ/mol/0.159 kJ/mol K

T = 1121 K

T (°C) = 1121 – 273 = 848

Standard Free Energies of FormationStandard Free Energies of Formation

G° = G°f (products) - G°f (reactants)

G° = [cG°f (C) + dG°f (D)] - [aG°f (A) + bG°f (B)]

ReactantsProducts

cC + dDaA + bB

Standard free energy of formation (G0f) is the free-energy

change that occurs when1 mole of the compound is formed from its elements in their standard ( 1 atm) states.

aA + bB cC + dD

G0rxn dG0 (D)fcG0 (C)f= [ + ] - bG0 (B)faG0 (A)f[ + ]

G0rxn nG0 (products)f= mG0 (reactants)f-

The standard free-energy of reaction (G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.

rxn

G0 of any element in its stable form is zero.

f --

Calculate G° for the equation below using the Gibbs free energy data given:

2 SO2(g) + O2(g) 2 SO3(g)

Gf° values (kJ): SO2(g) = 300.4, SO3(g) = 370.4

1. +70 kJ

2. +140 kJ

3. 140 kJ

4. 70 kJ

Correct Answer:

°° ° reactantsproducts ff GGG

G° = (2 370.4) [(2 300.4) + 0]

G° = 740.8 [600.8] = 140

1. +70 kJ

2. +140 kJ

3. 140 kJ

4. 70 kJ

Gibbs Free Energy07Gibbs Free Energy07

Calculation of Nonstandard ∆GCalculation of Nonstandard ∆G

• The sign of ∆G° tells the direction of spontaneous reaction when both reactants and products are present at standard state conditions.

• Under nonstandard( condition where pressure is not 1atm or concentrations of solutions are not 1M) conditions, ∆G˚ becomes ∆G.

∆G = ∆G˚ + RT lnQ

• The reaction quotient is obtained in the same way as an equilibrium expression.

Free Energy Changes and the Reaction MixtureFree Energy Changes and the Reaction Mixture

C2H4(g)2C(s) + 2H2(g) Qp =C2H4

P

H2P

2

Calculate ln Qp:

Calculate G for the formation of ethylene (C2H4) from carbon and hydrogen at 25 °C when the partial pressures are 100 atm H2 and 0.10 atm C2H4.

= -11.511002

0.10ln

G = G° + RT ln Q

Free Energy Changes and the Reaction MixtureFree Energy Changes and the Reaction Mixture

Calculate G:

= 68.1 kJ/mol +

G = 39.6 kJ/mol

(298 K)(-11.51)8.314K mol

J

G = G° + RT ln Q

1000 J

1 kJ

( see page A-13 of your book, page 667)

Free Energy and ChemicalEquilibrium 05Free Energy and ChemicalEquilibrium 05

• At equilibrium ∆G = 0 and Q = K

•∆G = ∆G˚ + RT lnQ

∆G˚rxn = –RT ln K

Free Energy and Chemical EquilibriumFree Energy and Chemical Equilibrium

Calculate Kp at 25 °C for the following reaction:

Calculate G°:

CaO(s) + CO2(g)CaCO3(s)

- [(1 mol)(-1128.8 kJ/mol)]

G° = +130.4 kJ/mol

= [(1 mol)(-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)]

G° = [G°f (CaO(s)) + G°f (CO2(g))] - [G°f (CaCO3(s))] Please see appendix B:

Free Energy and Chemical EquilibriumFree Energy and Chemical Equilibrium

Calculate K:

=

G° = -RT ln K

-130.4 kJ/mol

ln K = -52.63

(298 K)1000 J

1 kJ8.314

K mol

J

= 1.4 x 10-23

Calculate ln K:

RT

-G°ln K =

-52.63

K = e

Free Energy and ChemicalEquilibrium 04Free Energy and ChemicalEquilibrium 04

∆G˚rxn = –RT ln K

Suppose G° is a large, positive value. What then will be the value of the equilibrium constant, K?

1. K = 02. K = 13. 0 < K < 14. K > 1

Correct Answer:

G° = RTlnK

Thus, large positive values of G° lead to large negative values of lnK. The value of K itself, then, is very small.

1. K = 02. K = 13. 0 < K < 14. K > 1

More thermo?More thermo? You betcha!You betcha!

Calculate K for the reactionCalculate K for the reaction

NN22OO44 --->2 NO --->2 NO22 ∆G∆Goorxnrxn = +4.8 kJ/mole = +4.8 kJ/mole ( see page A-11)( see page A-11)

ΔG° = RT lnK

∆∆GGoorxnrxn = +4800 J = - (8.31 J/mol.K)(298 K) ln K = +4800 J = - (8.31 J/mol.K)(298 K) ln K

∆∆GGoorxnrxn = - RT lnK = - RT lnK

1.94- = K)J/m.K)(298 (8.31

J/mole 4800- = ln K

Thermodynamics and Thermodynamics and KKeqeq

K = 0.14K = 0.14When ∆GWhen ∆Goo

rxnrxn > 0, then K < 1 > 0, then K < 1