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 ProcessesSpontaneous 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.

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 (ΔH): The heat change in a reaction or process at constant pressure; ΔH = ΔE + PΔV.

Entropy (S): The amount of molecular randomness in a system.

EnthalpyEnthalpy

= +40.7 kJΔHvap

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

H2O(l)H2O(s)

H2O(g)H2O(l)

2NO2(g)N2O4(g)

= +3.88 kJΔH°

= +6.01 kJΔHfusion

= -890.3 kJΔH°

= +57.1 kJΔH°

Endothermic:

Exothermic:

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

EntropyEntropyΔS = Sfinal - Sinitial

EntropyEntropy

EntropyEntropy

The sign of entropy change, ΔS, 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; ΔSmust be positive.

1. Positive

2. Negative

3. Zero

Entropy and Temperature 02Entropy and Temperature 02

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

Ssolid < Sliquid < Sgas

Entropy Changes in the System (ΔSsys)

When gases are produced (or consumed)

• If a reaction produces more gas molecules than it consumes, ΔS0 > 0.

• If the total number of gas molecules diminishes, ΔS0 < 0.

• If there is no net change in the total number of gas molecules, then ΔS0 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, ΔS is negative.

Entropy and ProbabilityEntropy and Probability

S = k ln W

k = Boltzmann’s constant

= 1.38 x 10-23 J/K

W = The number of ways that the state can be achieved.

Standard Molar Entropies and Standard Entropies of ReactionStandard Molar Entropies and Standard Entropies of ReactionStandard 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 ΔH° for a reaction:

ΔH° = ΔH°f (Products) – ΔH°f (Reactants)

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

aA + bB cC + dDΔH° = [cΔH°f (C) + dΔH°f (D)] – [aΔH°f (A) + bΔH°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 (ΔSsys)

aA + bB cC + dD

ΔS0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

ΔS0rxn nS0

(products)= Σ mS0 (reactants)Σ-

The standard entropy change of reaction (ΔS0 ) 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

ΔS0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]

ΔS0rxn = 427.2 – [395.8 + 205.0] = -173.3J/K•mol

Consider 2 HConsider 2 H22(g) + O(g) + O22(g) (g) ------> 2 H> 2 H22O(liq)O(liq)∆∆SSoo = 2 S= 2 Soo (H(H22O) O) -- [2 S[2 Soo (H(H22) + S) + Soo (O(O22)])]∆∆SSoo = 2 mol (69.9 J/= 2 mol (69.9 J/KK••molmol) ) --

[2 mol (130.7 J/[2 mol (130.7 J/KK••molmol) + ) + 1 mol (205.3 J/1 mol (205.3 J/KK••molmol)])]

∆∆SSoo = = --326.9 J/K326.9 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 Calculating ∆∆S for a ReactionS for a Reaction

∆So = Σ So (products) - Σ So (reactants)∆∆SSoo = = ΣΣ SSoo (products) (products) -- ΣΣ SSoo (reactants)(reactants)

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

2 NO(g) + O2(g) → 2 NO2(g)ΔS° values (J/mol-K): NO2(g) = 240, NO(g) = 211, O2(g) = 205.

1. +176 J/mol2. +147 J/mol

3. −147 J/mol4. −76 J/mol

Correct Answer:

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

ΔS° = − 147

( ) ( )ΣΣ °° Δ−Δ=°Δ reactantsproducts SSS

1. +176 J/K2. +147 J/K

3. −147 J/K4. −76 J/K

Standard Molar Entropies 01Standard Molar Entropies 01

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 (ΔSsurr)

Exothermic ProcessΔSsurr > 0

Endothermic ProcessΔSsurr < 0

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

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

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

Δssurr α - ΔH

Δssurr α T1

Δssurr = T

- ΔH

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

2 H2 H22(g) + O(g) + O22(g) (g) ------> 2 H> 2 H22O(liq) O(liq) ∆∆H H °° = = --571.7 kJ571.7 kJ

∆∆SSoosystemsystem = = --326.9 J/K326.9 J/K

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

298.15 K

2nd Law of Thermodynamics2nd Law of Thermodynamics

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

ΔSsurr =−Δ 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

ΔS sur =−Δ Hsys

T•∆Stotal = ∆Ssys + ( ) > 0−ΔH sys

T

-T (∆Stotal = ∆Ssys + ( ) ) < 0−Δ H sys

T

Gibbs Free Energy 02Gibbs Free Energy 02

• 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

Calculate ∆Go rxn for the following:Calculate Calculate ∆∆GGo o rxnrxn for the following:for the following:CC22HH22(g) + 5/2 O(g) + 5/2 O22(g) (g) ----> 2 CO> 2 CO22(g) + H(g) + H22O(g)O(g)

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

rxnrxn = = --1238 kJ1238 kJUse standard molar entropies to calculate Use standard molar entropies to calculate ∆∆SSoo

rxnrxn ( see page ( see page 658)658)∆∆SSoo

rxnrxn = = --97.4 J/K or 97.4 J/K or --0.0974 kJ/K0.0974 kJ/K

∆∆GGoorxnrxn = = --1238 kJ 1238 kJ -- (298 K)((298 K)(--0.0974 kJ/K)0.0974 kJ/K)

= = --1209 kJ1209 kJ

Reaction is Reaction is productproduct--favoredfavored in spite of negative in spite of negative ∆∆SSoorxnrxn. .

Reaction is Reaction is ““enthalpy drivenenthalpy driven””

Calculate ΔG° for the equation below using the thermodynamic data given:

N2(g) + 3 H2(g) → 2 NH3(g)ΔH° (NH3) = −46 kJ; S° values (J/mol-K) are

NH3 = 192.5, N2 = 191.5, H2 = 130.6.

1. +33 kJ2. +66 kJ

3. −66 kJ4. −33 kJ

Correct Answer:

ΔH° = −92 kJΔS° = −198.7J/mol . K

ΔG° = ΔH° − TΔS°ΔG° = (−92) − (298)(−0.1987)ΔG° = (−92) + (59.2) = −33 KJ

1. +33 kJ2. +66 kJ3. −66 kJ4. −33 kJ

Gibbs Free Energy 03Gibbs Free Energy 03

Using ∆G = ∆H – T∆S, we can predict the sign of ∆Gfrom 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

Gibbs Free Energy 04Gibbs Free Energy 04

Gibbs Free Energy 06Gibbs Free Energy 06

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

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:

ΔS° = -198.7 J/K

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

= -33.0 kJ1000 J

ΔH° = -92.2 kJ

ΔG° = ΔH° - TΔS°

1 kJK

-198.7 Jx x= -92.2 kJ - 298 K

Gibbs Free Energy 05Gibbs Free Energy 05

• 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 ΔH° = 178.3 kJ/mol and ΔS° = 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 = ΔH°/ΔS°T = 178.3 kJ/mol/0.159 kJ/mol • K

T = 1121 KT (°C) = 1121 – 273 = 848

Standard Free Energies of FormationStandard Free Energies of Formation

ΔG° = ΔG°f (products) - ΔG°f (reactants)

ΔG° = [cΔG°f (C) + dΔG°f (D)] - [aΔG°f (A) + bΔG°f (B)]

ReactantsProducts

cC + dDaA + bB

Gibbs Free Energy 07Gibbs Free Energy 07

Standard free energy of formation (ΔG0f) is the free-energy

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

f

aA + bB cC + dDΔG0

rxn dΔG0 (D)fcΔG0 (C)f= [ + ] - bΔG0 (B)faΔG0 (A)f[ + ]

ΔG0rxn nΔG0 (products)f= Σ mΔG0 (reactants)fΣ-

The standard free-energy of reaction (ΔG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.

rxn

ΔG0 of any element in its stable form is zero.

f--

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

2 SO2(g) + O2(g) 2 SO3(g)ΔG° values (kJ): SO2(g) = −300.4, SO3(g) = −370.4

1. +70 kJ2. +140 kJ3. −140 kJ4. −70 kJ

Correct Answer:

( ) ( )∑ ∑°°

−=° reactantsproducts ff GGG ΔΔΔ

ΔG° = (2 × −370.4) − [(2 × −300.4) + 0]ΔG° = −740.8 − [−600.8] = −140

1. +70 kJ2. +140 kJ3. −140 kJ4. −70 kJ

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 ΔG 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.

Calculate ln Qp:

= -11.511002

0.10ln

ΔG = ΔG° + RT ln Q

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

Calculate ΔG:

= 68.1 kJ/mol + 8.314K mol

J

ΔG = 39.6 kJ/mol

(298 K)(-11.51)

ΔG = ΔG° + RT ln Q

1000 J1 kJ

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 ΔG°:

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

- [(1 mol)(-1128.8.0 kJ/mol)]

ΔG° = +130.4 kJ/mol

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

ΔG° = [ΔG°f (CaO(s)) + ΔG°f (CO2(g))] - [ΔG°f (CaCO3(s))]

Free Energy and Chemical EquilibriumFree Energy and Chemical Equilibrium

Calculate K:

=

ΔG° = -RT ln K

-130.4 kJ/mol

ln K = -52.63

(298 K)1000 J

1 kJ8.314

K molJ

= 1.4 x 10-23

Calculate ln K:

RT-ΔG°

ln K =

-52.63K = e

Free Energy and ChemicalEquilibrium 04Free Energy and ChemicalEquilibrium 04∆G˚rxn = –RT ln K

Suppose ΔG° 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:

ΔG° = −RTlnK

Thus, large positive values of ΔG°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 You betchabetcha!!

Calculate K for the reactionCalculate K for the reactionNN22OO44 ------>2 NO>2 NO22 ∆∆GGoo

rxnrxn = +4.8 kJ/mole = +4.8 kJ/mole ( see page ( see page

667)667)

ΔG° = −RTlnK∆∆GGoo

rxnrxn = +4800 J = = +4800 J = -- (8.31 J/mol.K)(298 K) (8.31 J/mol.K)(298 K) lnln KK

∆∆GGoorxnrxn = = -- RT RT lnKlnK

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

J/mole 4800- = ln K

Thermodynamics and Thermodynamics and KKeqeq

K = 0.14K = 0.14When When ∆∆GGoo

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