Chapter 17 1. The Austrian physicist Ludwig Boltzmann introduced a model to relate the total number...

1

Chapter 17

The Austrian physicist Ludwig Boltzmann introduced a model to relate the total number of microstates (the multiplicity, W) to entropy (S).

S = k ln W

AN

Rk

Boltzmann constant (k)

units on S is J/K

k = 1.38 x 1023J/K

Gas constant

Avogadro constant

Microstates and Entropy

Boltzmann was the founding father of statistical mechanics, a completely new way of thinking about theoretical physics. His work was ridiculed by his fellow professors.

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Standard Entropy

• Difficult to count the number of microstates directly.

• Measured by calorimetery.

• Standard entropy (S) absolute entropy of a substance at 1 atm (typically at 25C)

• Units of J K-1 mol-1

• All positive values.

• Absolute values (unlike DH which is relative)

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Why is there a –S?

S = k ln W

Al + 6H2O Al(H2O)6 (aq)3+ 2+

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Chapter 17

6

Spontaneity

Spontaneous: process that does occur under a specific set of conditions.

Nonspontaneous: process that does not occur under a specific set of conditions.

The reverse of a spontaneous process is a nonspontaneous one.

Spontaneity depends on temperature.

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Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

Predicting Spontaneity

DHrxn = Hproducts – Hreactants

Hproducts < Hreactants

DH < 0

Hproducts > Hreactants

DH > 0Exothermic Endothermic

Enthalpically favorable Enthalpically unfavorable

If we mix reactants and products together will the reaction occur?

Not enough information. We need to know the change in entropy to predict if a reaction will spontaneously occur.

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Entropy (S) Is a measure of the number of specific ways (microstates) in which a thermodynamic system may be arranged.


DSrxn = Sproducts – Sreactants

Sproducts < Sreactants

DS < 0

Sproducts > Sreactants

DS > 0Entropy decreases Entropy increases

Entropically unfavorable Entropically favorable


Still not enough information. We need to know total entropy change to predict if a reaction will spontaneously occur.

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Nonspontaneous process:

DSuniv = DSsys + DSsurr

Equilibrium : DSuniv = 0

DSuniv < 0

Spontaneous process: DSuniv > 0


Can calculate from:

DS0rxn nS0(products)= S mS0(reactants)S-

If we know DSsurr we can calculate DSuniv

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Can calculate from:

DS0rxn nS0(products)= S mS0(reactants)S-

If we know DSsurr we can calculate DSuniv


DS0 rx

n

surr

ound

ing

syst

em

DS su

rr

Surrounding = everything in the universe except the system.

Very, very difficult to measure!If not impossible.

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Exothermic Reaction Endothermic Process

Heat and Entropy

DSsurr > 0

-DHsys

Surroundings + heat = ↑ S

0 < DHsys

DSsurr < 0

+DHsys

Surroundings - heat = ↓ S

0 > DHsys

DSsurr ∝ -DHsys

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Heat and EntropyHeat released by the system increases the disorder of the surroundings.

DSsurr ∝ -DHsys

The effect of -DHsys on the surroundings depends on temperature: – At high temperature, where there is already considerable

disorder, the effect is muted

– At low temperature the effect is much more significant

– The difference between tossing a rock into a calm pool (low T) and a storm-tossed ocean (high T)

DSsurr = -DHsys

T

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DSsurr = -DHsys

TDSuniv = DSsys + -DHsys

T

-TDSuniv = -TDSsys + DHsys

Substitution:

Multiply by -T:

-TDSuniv = DHsys - TDSsysRearrange:

Both in terms of the system.

This equation relates DSuniv to DHsys and DSsys.

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Gibbs Free Energy

-TDSuniv = DHsys - TDSsys

Josiah Willard Gibbs (1839-1903)

DG = DHsys - TDSsys

First American Ph.D. in Engineering (Yale, 1863)

Praised by Albert Einstein as "the greatest mind in American history"

Died at 64 from an acute intestinal obstruction.

Gibbs free energy (DG)-

• AKA Free energy.

• Relates S, H and T of a system.

• Can be used to predict spontaneity.

• G is a state function.

in K

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Gibbs Free Energy

DG = DHsys - TDSsys

Gibbs free energy (DG)- Can be used to predict spontaneity.

For a constant temperature and constant pressure process:

DG < 0 The reaction is spontaneous in the forward direction.

DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.

DG = 0 The reaction is at equilibrium.

-DG = -T(+DSuniv)

+DG = -T(-DSuniv)

DSuniv > 0

-TDSuniv = DHsys - TDSsys

DSuniv < 0

DG = -T(DSuniv) = 0 DSuniv = 0

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Gibbs Free Energy

DG = DH - TDS• If you know DG for reactants and products then you can

calculate if a reaction is spontaneous. • If you know DG for two reaction then you can calculate if the

sum is spontaneous. • If you know DS, DH and T then you can calculate spontaneity. • Can predict the temperature when a reaction becomes

spontaneous.• If you have DHvap or DHfus and DS you can predict boiling and

freezing points.• If you have DHvap or DHfus and T you can predict the entropy

change during a phase change.• Can predict equilibrium shifts.

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Standard Free Energy ChangesThe standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.

rxn

aA + bB cC + dD

DG0rxn nDG0 (products)f= S mDG0 (reactants)fS-

DG0rxn dDG0 (D)fcDG0 (C)f= [ + ] - bDG0 (B)faDG0 (A)f[ + ]

Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states.

f

DG0 of any element in its stable form is zero.f

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aA + bB cC + dD

Standard Free Energy ChangesDG is a state function so free energy can be calculated from the

table of standard values just as enthalpy and entropy changes.

DG0rxn nDG0 (products)f= S

mDG0 (reactants)fS-

Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states.

f

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Standard Free Energy Changes

From appendix 3:KClO3(s) DGf = -289.9 kJ/molKCl(s) DGf = -408.3 kJ/molO2(g) DGf = 0 kJ/mol

2KClO3(s) 2KCl(s) + 3O2(g)

Calculate the standard free-energy change for the following reaction:

DG0rxn nDG0 (products)f= S mDG0 (reactants)fS-

DG0rxn = [2(408.3 kJ/mol) + 3(0)] [2(289.9 kJ/mol)]

DG0rxn = 816.6 (579.8)

DG0rxn = 236.8 kJ/mol

DG0rxn < 0 Yes!

Is the reaction spontaneous?

Example

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17.4

Calculate the standard free-energy changes for the following reactions at 25°C.

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

From appendix 3:CH4(g) DGf = -50.8 kJ/molO2(g) DGf = 0 kJ/molCO2(g) DGf = -394.4 kJ/molH2O (l) DGf = -237.2 kJ/mol

ΔG°rxn = [ΔG°f (CO2) + 2ΔG°f (H2O)] - [ΔG°f (CH4) + 2ΔG°f (O2)]

ΔG°rxn =[(-394.4 kJ/mol) + (2)(-237.2 kJ/mol)] - [(-50.8 kJ/mol) + (2) (0 kJ/mol)]

ΔG°rxn = -818.0 kJ/mol

Spontaneous?

Yes.

Example

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17.4

Calculate the standard free-energy changes for the following reactions at 25°C.

(b) 2MgO(s) 2Mg(s) + O2(g)

From appendix 3:MgO(s) DGf = -569.6 kJ/molO2(g) DGf = 0 kJ/molMg(s) DGf = 0 kJ/mol

ΔG°rxn = [2ΔG°f (Mg) + ΔG°f (O2)] - [2ΔG°f (MgO)]

ΔG°rxn = [(2)(0 kJ/mol) + (0 kJ/mol)] - [(2)(-569.6 kJ/mol)]

ΔG°rxn = 1139 kJ/mol

Spontaneous?

No.But the reverse is…

2Mg(s) + O2(g) 2MgO(s)

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More DG° Calculations

DG0rxn nDG0 (products)

f= S

mDG0 (reactants)fS-

To predict spontaneity of any rxn:

1) Pick any reactants and products.

2) Write a balanced equation.

3) Calculate DG0rxn.

4) Is the reaction spontaneous?

Appendix 3

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Another Example

+ O2

C(s, diamond) + O2(g) CO2(g)

DG°rxn = -396.4 kJ

Therefore, diamonds are contributing to global warming!

very slowly

ΔG°rxn = [ΔG°f (CO2)] - [ΔG°f (C, diamond) + ΔG°f (O2)]

From appendix 3:C, diamond(s) DGf = 2.9 kJ/molO2(g) DGf = 0 kJ/molCO2(g) DGf = -394.4 kJ/mol

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More DG° CalculationsSimilar to DH°, one can use the DG° for various reactions to determine DG° for the reaction of interest (a “Hess’ Law” for DG°) Hess’ Law- states that regardless

of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes.

What is the DG° for this reaction:

Given:C(s, diamond) + O2(g) CO2(g) DG° = -397 kJ

C(s, graphite) + O2(g) CO2(g) DG° = -394 kJ

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More DG° CalculationsWhat is the DG° for this reaction:

C(s, graphite) + O2(g) CO2(g) DG° = -394 kJ

CO2(g) C(s, graphite) + O2(g) DG° = +394 kJ

Given:C(s, diamond) + O2(g) CO2(g) DG° = -397 kJ

C(s, diamond) + O2(g) CO2(g) DG° = -397 kJ

CO2(g) C(s, graphite) + O2(g) DG° = +394 kJ

C(s, diamond) C(s, graphite) DG° = -3 kJ

DG°rxn < 0…..rxn is spontaneous

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Alternative DG CalculationIs the following reaction spontaneous at 298 K?

(assume standard conditions)

KCl(s)(s)3KClO(s)4KClO 43

DH°f (kJ/mol) S° (J/mol.K)

KClO3(s) -397.7 143.1

KClO4(s) -432.8 151.0

KCl (s) -436.7 82.6

Given:

DG°rxn = DH°rxn - TDS°rxn

First Calculate

Given

Then Find

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Alternative DG CalculationKCl(s)(s)3KClO(s)4KClO 43

DH°f (kJ/mol) S° (J/mol.K)KClO3(s) -397.7 143.1KClO4(s) -432.8 151.0KCl (s) -436.7 82.6

kJ

kJkJkJ

KClOHKClHKClOHH fffrxn

144

)7.397(4)7.436()8.432(3

43 34

KJ

KJ

KJ

KJ

KClOSKClSKClOSS rxn

8.36

)1.143(4)6.82()0.151(3

43 34

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Alternative DG CalculationKCl(s)(s)3KClO(s)4KClO 43

DH°f (kJ/mol) S° (J/mol.K)KClO3(s) -397.7 143.1KClO4(s) -432.8 151.0KCl (s) -436.7 82.6

DH°rxn = -144 kJ DS°rxn = -36.8 J/K

G rxn H rxn TS rxn

144kJ 298K 38.6J K 1kJ

1000J

133kJ

DG°rxn < 0…..rxn is spontaneous at 298 K

Enthalpically favorable Entropically unfavorable

What about at 5000 K?

DG°rxn = 50 kJDG°rxn > 0…rxn is nonspontaneous at 5000 K

(Assuming the same DH and S)

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DG Temperature Dependence


DG°rxn = 50 kJ at 5000 K

DG°rxn = -133 kJ at 298 K Spontaneous

Nonspontaneous

Reaction spontaneity is a temperature dependent phenomenon!

DGrxn = DHrxn - TDSrxn

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DG = DH - TDS

Enthalpy:DHrxn < 0 The reaction is enthalpically favorable.

Entropy: DSrxn > 0 The reaction is entropically favorable.

Need both to predict spontaneity. And sometimes temperature!DG < 0 SpontaneousDG > 0 NonspontaneousDG = 0 Equilibrium

1) If DH < 0 and DS > 0, then DG is negative at all T

3) If DH < 0 and DS < 0, then DG depends on T

2) If DH > 0 and DS > 0, then DG depends on T

4) If DH > 0 and DS < 0, then DG is positive at all T

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DG = DH - TDSDG < 0 SpontaneousDG > 0 NonspontaneousDG = 0 Equilibrium

3) DH < 0 and DS < 0

If DH < TDS, then DG is positive.If DH > TDS, then DG is negative.


2) DH > 0 and DS > 0

If DH < TDS, then DG is negative.If DH > TDS, then DG is positive.

Nonspontaneous at high TSpontaneous at low T

(Enthalpically favorable, entropically unfavorable)

(Enthalpically unfavorable, entropically favorable)

Spontaneous at high TNonspontaneous at low T

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DG = DH - TDS


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Nonspontaneous

Spontaneous

Four possible scenarios:

(1) H < 0, S > 0

(2) H > 0, S > 0

(3) H < 0, S < 0

(4) H > 0, S < 0

exothermic

system becomes more disordered

exothermic

system becomes more ordered

endothermic

system becomes more disordered

DSuniv and Spontaneous ReactionsDG = DH - TDS

Class 1: Class 3:

Class 2:

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Predicting T from Gibbs Equation


DG°rxn = 50 kJ at 5000 K

DG°rxn = -133 kJ at 298 K Spontaneous

Nonspontaneous

Reaction spontaneity is a temperature dependent phenomenon!

At what temperature will the reaction become spontaneous?

Find T where DG changes from positive to negative.

I.e. when DG =0.

DG = DH – TDS = 0

T = DH/DS

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Predicting T from Gibbs EquationAt what T is the following reaction spontaneous?

Br2(l) Br2(g)

Given: DH°= 30.91 kJ/mol

DS°= 93.2 J/mol.K

DG = DH – TDS = 0

T = DH/DS

T = (30.91 kJ/mol) /(93.2 J/mol.K)

T = 331.7 K

H > 0, S > 0The reaction will be spontaneous when

T > 331.7 K

Predicting T from Gibbs EquationAt what T is the following reaction spontaneous?

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

DG0 = DH0 – TDS0 = 0

T = D0H/DS0

DH0 = 177.8 kJ/mol

DS0 = 160.5 J/K·mol

DG0 = 0 at 835 oC

Equilibrium Pressure of CO2

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Chapter 11

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Liquid ↔ Gas

At time = 0 At time > 0 At time = ∞

Liquid

# of molecules in =# of molecules out

At equilibrium!

rate of evaporation = rate of condensation

DG = DH - TDSDG < 0 SpontaneousDG > 0 NonspontaneousDG = 0 Equilibrium

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Gibbs Equation and Phase Change

Molar heat of vaporization (DHvap) is the energy required to vaporize 1 mole of a liquid at its boiling point.

H2O (l) H2O (g)

DG = DH – TDS = 0

If we know DHvap and boiling point we can calculate DS!

DS = TDH

= 40.79 kJ/mol

373 KDS

DHvap = 40.79 kJ/molBP(H2O) = 373 K

Example

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17.5

The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid → liquid and liquid → vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.

solid

liquid

vapor

Example

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17.5


Molar heat of vaporization (DHvap) is the energy required to vaporize 1 mole of liquid to gas.

Molar heat of fusion (DHfus) is the energy required to melt 1 mole of solid.

solid → liquid → vaporDHvapDHfus

Example

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17.5


solid → liquid → vaporDHvapDHfus

DG = DH – TDS = 0 DS = TDH

fusfus

f

ΔΔ =

(10.9 kJ/mol)(1000 J/1 kJ) =

(5.5 + 273)K

= 39.1 J / K mol

HS

Tvap

vapb

ΔΔ =

(31.0 kJ/mol)(1000 J/1 kJ) =

(80.1 + 273)K

= 87.8 J / K mol

HS

T

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Chapter 17

Chapter 17 1. The Austrian physicist Ludwig Boltzmann introduced a model to relate the total number...

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