1

Chapter 17

2

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

3

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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Chapter 17

5

The 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-

Free Energy and Equilibrium

DG° < 0 favors products spontaneouslyDG° > 0 favors reactants spontaneously

Does not tell you it will go to completion!

aA + bB cC + dDDG° fwd

DG° rev

DG° fwd = -DG° rev

The value of G° calculated under the standard conditions characterizes the “driving force” of the reaction towards equilibrium.

6

DG° = -R T lnK

Free Energy and Equilibrium

standard free-energy(kJ/mol) gas constant

(8.314 J/Kmol)

temperature(K)

equilibrium constant(Kp, Kc, Ka, Ksp, etc.)

• Arguably most important equation in chemical thermodynamics!

• It allows us to calculate the extent of a chemical reaction if its enthalpy and entropy changes are known.

• The changes in enthalpy and entropy can be evaluated by measuring the variation of the equilibrium constant with temperature.

• This relationship is only valid for the standard conditions, i.e. when the activities of all reactants and products are equal to 1.

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Derivationhttp://en.wikipedia.org/wiki/Chemical_equilibrium#Thermodynamics

8

9

FO

RW

AR

D R

EA

CT

ION

RE

VE

RS

E R

EA

CT

ION

DG0(kJ) K Significance

200

100

50

10

1

0

-1

-10

-50

-100

-200

9x10-36

3x10-18

2x10-9

2x10-2

7x10-1

1

1.5

5x101

6x108

3x1017

1x1035

Essentially no forward reaction; reverse reaction goes to completion

Forward and reverse reactions proceed to same extent

Forward reaction goes to completion; essentially no reverse reaction

Free Energy and EquilibriumDG° = -R T lnK

R is constant so at a given temperature:

10

Using DG° and KUsing the table of standard free energies, calculate the equilibrium constant, KP, for the following reaction at 25 C.

2HCl(g) H2(g) + Cl2(g)

From appendix 3:H2(g) DGf = 0 kJ/molCl2(s) DGf = 0 kJ/molHCl(g) DGf = -95.3 kJ/mol

DG0rxn

nDG0 (products)f= S mDG0 (reactants)

fS-

DG0rxn = [1(0) + 1(0)] - [2(95.3 kJ/mol)]

DG0rxn = 190.6 kJ/mol Non-spontaneous!

Favors reactants!

DG° = -R T ln K

11

Using DG° and KUsing the table of standard free energies, calculate the equilibrium constant, KP, for the following reaction at 25 C.

2HCl(g) H2(g) + Cl2(g)

DG0rxn

nDG0 (products)f= S mDG0 (reactants)

fS-

DG0rxn = [1(0) + 1(0)] - [2(95.3 kJ/mol)]

DG0rxn = 190.6 kJ/mol Non-spontaneous!

Favors reactants!

DG° = -R T ln K190.6 kJ/mol = (8.314 J/K·mol)(25C) ln KP

correct units

190.6 kJ/mol = (8.314 x 103 kJ/K·mol)(298 K) ln KP

KP = 3.98 x 1034 Favors reactants!

Example

12

17.6

Using data listed in Appendix 3, calculate the equilibrium constant (KP) for the following reaction at 25°C:

2H2O(l) 2H2(g) + O2(g)

ΔG°rxn = [2ΔG°f(H2) + ΔG°f(O2)] - [2ΔG°f(H2O)]

= [(2)(0 kJ/mol) + (0 kJ/mol)] - [(2)(-237.2 kJ/mol)]

= 474.4 kJ/molorxn p

p

p

p

Δ = - ln

1000 J474.4 kJ/mol × = -(8.314J/K mol)(298 K)ln

1 kJ ln = -191.5

= . -847 × 10

G RT K

K

K

K e 1915

Bonus: What is the Kp for the reverse reaction?

2H2(g) + O2(g) 2H2O(l)

ΔG°rxn = -474.4 kJ/mol or 1/Kp(fwd) = Kp(rev)

lnKp(fwd) = -lnKp(rev)

Example

13

17.7

In Chapter 16 we discussed the solubility product of slightly soluble substances. Using the solubility product of silver chloride at 25°C (1.6 x 10-10), calculate ΔG° for the process

AgCl(s) Ag+(aq) + Cl-(aq)

Ksp = [Ag+][Cl-] = 1.6 x 10-10

ΔG° = -(8.314 J/K·mol) (298 K) ln (1.6 x 10-10)

= 5.6 x 104 J/mol

= 56 kJ/mol

Favors reactants.Not very soluble!

DG° = -R T lnK

Free Energy and EquilibriumDG° = -R T lnK

ln K = - DG° R T

Substitution:

Rearrange:

DG = DH - TDSln K = -DH - TDS

R T

ln K = -DH TDS R T R T

Rearrange: +

( )ln K = -DHR R

+DS

T1

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( )

Free Energy and EquilibriumDG° = -R T lnK

ln K = - DG° R T

Substitution:

Rearrange:

DG = DH - TDSln K = -DH - TDS

R T

ln K = -DH TDS R T R T

Rearrange:

ln K = -DHR R

+

+DS

T1

y = m • x + b

Measure equilibrium with respect to temperature:

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Free Energy and Equilibrium

rateAB = kobs [A]

rateBA = kf [B]

A B

At equilibrium:

rateAB = rateBA

kobs [A] = kf [B]

[B]=

kobs

[A]kf

= Ku

kobs

kf

Find the DS and DH of the following:

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Free Energy and Equilibrium

kobs

kf

Slope = -7261.1 K

DH° = 60 kJ/molDS° = 200 J/KmolR

DS

-DHR

Find the DS and DH of the following:

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DG° vs DGYou have already delta ΔG°, ΔS° and ΔH° in which the ° indicates that all components are in their standard states.

Definition of “standard”:

*There is no "standard temperature", but we usually use 298.15 K (25° C).

• Even if we start a reaction at standard conditions (1 M) the reaction will quickly deviate from standard.

• DG° indicates whether reactants or products are favored at equilibrium.

• DG at any give time is used to predict the direction shift to reach equilibrium.

• If a mixture is not at equilibrium, the liberation of the excess Gibbs free energy (DG) is the “driving force” for the composition of the mixture to change until equilibrium is reached. 18

Free Energy and Equilibrium

DG° = -R T ln K

standard free-energy(kJ/mol) gas constant

(8.314 J/Kmol)

temperature(K)

reaction quotient

At equilibrium:

DG = DG° + R T ln QAt any time:

non-standard free-energy(kJ/mol)

The sign of DG tells us that the reaction would have to shift to the left to reach equilibrium.

DG < 0, reaction will shift rightDG > 0, reaction will shift leftDG = 0, the reaction is at equilibrium

The magnitude of DG tells us how far it has to go to reach equilibrium. 19

If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (DG < 0)

If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (DG > 0)

If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (DG = 0)

Free Energy and Equilibrium

standard free-energy(kJ/mol) gas constant

(8.314 J/Kmol)

temperature(K)

reaction quotientDG = DG° + R T ln Q

non-standard free-energy(kJ/mol)

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21

Another Example

H2(g) + Cl2(g) 2 HCl(g)

For the following reaction at 298 K:

H2 = 0.25 atmCl2 = 0.45 atmHCl = 0.30 atm

Given: From appendix 3:

H2(g) DGf = 0 kJ/molCl2(s) DGf = 0 kJ/molHCl(g) DGf = -95.3 kJ/mol

Which way will the reaction shift to reach equilibrium?

DG = DG° + R T ln Q

constantgiven

calculatecalculate

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Another Example

H2(g) + Cl2(g) 2 HCl(g)

For the following reaction at 298 K:

H2 = 0.25 atmCl2 = 0.45 atmHCl = 0.30 atm

Given: From appendix 3:

H2(g) DGf = 0 kJ/molCl2(s) DGf = 0 kJ/molHCl(g) DGf = -95.3 kJ/mol

Which way will the reaction shift to reach equilibrium?

DG = DG° + R T ln Q

G° = [2(95.27 kJ/mol)] [0 + 0] = 190.54 kJ/mol

2 2( ) (0.30)HCl 0.80( ) ( ) (0.25) (0.45)

H Cl2 2

PQP P P

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Another Example

H2(g) + Cl2(g) 2 HCl(g)

For the following reaction at 298 K:

H2 = 0.25 atmCl2 = 0.45 atmHCl = 0.30 atm

Given: From appendix 3:

H2(g) DGf = 0 kJ/molCl2(s) DGf = 0 kJ/molHCl(g) DGf = -95.3 kJ/mol

Which way will the reaction shift to reach equilibrium?

DG = DG° + R T ln Q

G = 190,540 J/mol + (8.314J/K·mol)(298 K) ln (0.80)

G° = 190.54 kJ/mol Q = 0.80constantgiven

G = 191.09 kJ/mol Because ΔG < 0, the net reaction proceeds from left to right to reach equilibrium.

Example

24

17.8

The equilibrium constant (KP) for the reaction

N2O4(g) 2NO2(g)

is 0.113 at 298 K, which corresponds to a standard free-energy change of 5.40 kJ/mol. In a certain experiment, the initial pressures are PNO2

= 0.122

atm and PN2O4 = 0.453 atm.

Calculate ΔG for the reaction at these pressures and predict the direction of the net reaction toward equilibrium.

2

2 4

op

2NOo

N O

23

3 3

3

Δ = Δ + ln

= Δ + ln

(0.122) = 5.40 × 10 J/mol + (8.314J/K mol)(298 K) × ln

0.453

= 5.40 × 10 J/mol - 8.46 × 10 J/mol

= -3.06 × 10 J/mol = - 3.06kJ / mol

G G RT Q

PG RT

P

Because ΔG < 0, the net reaction proceeds from left to right to reach equilibrium.

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DG0 < 0 DG0 > 0

Free Energy and EquilibriumDG° = -R T ln KAt equilibrium:

DG = DG° + R T ln QAt any time:

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Chapter 17

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Synthesis of proteins: (first step) alanine + glycine alanylglycine G° = 29 kJ/mol

“Uphill” Reactions

Because ΔG > 0, the reaction is non-spontaneous.

alanine

glycineNo reaction!

Need to couple two reactions!

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Coupled ReactionsCoupled Reactions- using a thermodynamically favorable (G° < 0) reaction (G° < 0) to drive an unfavorable one (G° > 0) .

Example: Industrial ore separation-

Sphalerite ore

Major applications in the US1) Galvanizing (55%)2) Alloys (21%)3) Brass and bronze (16%)4) Miscellaneous (8%)

Zinc Metal

We need 2000 tones of the zinc metal per year!

White pigment (ZnO)Fire retardant (ZnCl2)Vitamin supplement (Zn2+)Reducing agent (Zn(s))

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Coupled ReactionsCoupled Reactions- using a thermodynamically favorable (G° < 0) reaction (G° < 0) to drive an unfavorable one (G° > 0) .

Example: Industrial ore separation-

95 % of Zinc is produced by this method

Unfavorable reaction (G° > 0)

Coupled Reactions in Biology

glucose + Pi → glucose-6-phosphate ATP + H2O → ADP + Piglucose + ATP → glucose-6-phosphate + ADP 30

G° < 0

G° > 0

Coupled Reactions in Biology

FoodStructural motion and maintenance

Fats and Carbohydrates

ATP and NADPH

Coupled reactions

?

Chemical Batteries for the BodyStored bond energy

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Coupled Reactions in BiologyDigestion/respiration:

Generation of ATP:

Burning Glucose

Low Energy Higher Energy32

Synthesis of proteins: (first step)

“Uphill” Reactions

ATP + H2O ADP + H3PO4 G° = -31 kJ/mol

alanine + glycine alanylglycine G° = 29 kJ/mol

alanine + glycine + ATP + H2O alanylglycine + ADP + H3PO4 G° = -2 kJ/mol

Spontaneous!

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34

Coupled Reactions in Biology

Coupled reactions to drive the synthesis of:AminoacidsRiboseNucleic acidsPolypeptidesDNAPhospholipids

This is why we eat!…and why plants absorb light.

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Chapter 17