Chapter 17Free Energy

and Thermodynamics

2

Thermodynamics vs. Kinetics

CHEMICAL THERMODYNAMICSCHEMICAL THERMODYNAMICS

The first law of thermodynamicsThe first law of thermodynamics::Energy and matter can be neither created nor Energy and matter can be neither created nor destroyed; only transformed from one form to destroyed; only transformed from one form to

another.another. The energyThe energy and matter of the universe and matter of the universe is constant.is constant.

The second law of thermodynamicsThe second law of thermodynamics::In any spontaneous process there is always an In any spontaneous process there is always an

increase in the entropy of the universe.increase in the entropy of the universe. The entropy is increasing.The entropy is increasing.

The third law of thermodynamicsThe third law of thermodynamics::The entropy of a perfect crystal at 0 K is zero.The entropy of a perfect crystal at 0 K is zero. There is no molecular motion at absolute 0 K.There is no molecular motion at absolute 0 K.

STATE FUNCTIONSSTATE FUNCTIONS

A property of a system which depends only on A property of a system which depends only on its present state and not on its pathway.its present state and not on its pathway.

H = Enthalpy = heat of reaction = qH = Enthalpy = heat of reaction = qpp A measure of heat (energy) flow of a system A measure of heat (energy) flow of a system relative to its surroundings.relative to its surroundings.

H° standard enthalpyH° standard enthalpyHHff° enthalpy of formation° enthalpy of formation

H° = H° = n n HHff° (products) - ° (products) - m m HHff° (reactants)° (reactants)

H = H = U + PU + PVVU represents the Internal energy of the U represents the Internal energy of the particles, both the kinetic and potential energy. particles, both the kinetic and potential energy. U = q + wU = q + w

HEAT VS WORK

energy transfer as a energy expanded to result of a temperature move an object againstdifference a force

qp w = F x d

endothermic (+q) work on a system(+w)

exothermic (-q) work by the system(-w)

qc = -qh w = -PV

SPONTANEOUS PROCESSESA spontaneous process occurs without outside intervention. The rate may be fast or slow.

EntropyA measure of randomness or disorder in a system.

Entropy is a state function with units of J/K and it can be created during a spontaneous process.

Suniv = Ssys + Ssurr

The relationship between Ssys and Ssurr

Ssys Ssurr Suniv Process spontaneous? ++ + + + + Yes -- - - - - No (Rx will occur in

opposite direction) ++ - - ? ? Yes, if Ssys > Ssurr

-- + + ? ? Yes, if Ssurr > Ssys

7

Comparing Potential Energy

The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.

8

Reversibility of Process• any spontaneous process is irreversible

– it will proceed in only one direction• a reversible process will proceed back and forth

between the two end conditions– equilibrium– results in no change in free energy

• if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction

9

Diamond → Graphite

Graphite is more stable than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in your lifetime (or through many of

your generations).

Predicting Relative S0 Values of a System

1. Temperature changes

2. Physical states and phase changes

3. Dissolution of a solid or liquid

5. Atomic size or molecular complexity

4. Dissolution of a gas

S0 increases as the temperature rises.

S0 increases as a more ordered phase changes to a less ordered phase.

S0 of a dissolved solid or dissolved liquid is usually greater than the S0 of the pure solute. However, the extent depends upon the nature of the solute and solvent.

A gas becomes more ordered when it dissolves in a liquid or solid.

In similar substances, increases in mass relate directly to entropy.

In allotropic substances, increases in complexity (e.g. bond flexibility) relate directly to entropy.

11

Factors Affecting Whether a Reaction Is Spontaneous

• The two factors that determine the thermodynamic favorability are the enthalpy and the entropy.

• The enthalpy is a comparison of the bond energy of the reactants to the products. – bond energy = amount needed to break a bond. – H

• The entropy factors relates to the randomness/orderliness of a system – S

• The enthalpy factor is generally more important than the entropy factor

Entropy Changes in the System

S0rxn - the entropy change that occurs when all reactants

and products are in their standard states.

S0rxn = S0

products - S0reactants

The change in entropy of the surroundings is directly related to an opposite change in the heat of the system and inversely related to the temperature at which the heat is transferred.

Ssurroundings = - Hsystem

T

Entropy

S = SS = Sff - S - Sii S > q/TS > q/TS = S = H/TH/T

For a reversible (at equilibrium) processH - T S < 0

For a spontaneous reaction at constant T & P H - T S

If the value for H - T S is negative for a reaction then the reaction is spontaneous in the

direction of the products.

If the value for H - T S is positive for a reaction then the reaction is spontaneous in the direction of the reactants. (nonspontaneous for products)

A spontaneous endothermic chemical reaction.

water

Ba(OH)2 8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l).

H0rxn = +62.3 kJ

APPLICATION OF THE 3RD LAW OF THERMODYNAMICS

S° = standard entropy = absolute entropy

S is usually positive (+) for Substances, S can be negative (-) for Ions because H3O+ is used as zero

Predicting the sign of S° The sign is positive if:1. Molecules are broken during the Rx2. The number of moles of gas

increases3. solid liquid liquid gas solid

gas an increase in order occurs

1. Ba(OH)2•8H2O + 2NH4NO3(s) 2NH3(g) + 10H2O(l) + Ba(NO3)2(aq)

2. 2SO(g) + O2(g) 2SO3(g)

3. HCl(g) + NH3(g) NH4Cl(s)

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

KJ3

KkJ

surr

syssurr

106.86 86.6S

K 298

kJ 2044

T

HS

Ex. 17.2a – The reaction C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) has Hrxn = -2044 kJ at 25°C.

Calculate the entropy change of the surroundings.

combustion is largely exothermic, so the entropy of the surrounding should increase significantly

Hsystem = -2044 kJ, T = 298 K

Ssurroundings, J/K

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

ST, H

T

HS sys

surr

S°= n S°(product)- m S°(reactant)

1. Acetone, CH3COCH3, is a volitale liquid solvent. The standard enthalpy of formation of the liquid at 25°C is -247.6 kJ/mol; the same quantity for the vapor is -216.6 kJ/mol. What is S when 1.00 mol liquid acetone vaproizes?

2.Calculate S° at 25° for:

a. 2 NaHCO3 (s) CO2 (g) + Na2CO3 (s) + H2O (g)

b. 2 AgBr(s) + I2(s) Br2(l) + 2 AgI(s)

Table of Sf

G0system = H0

system - TS0system

G0rxn = mG0

products - nG0reactants

Gibbs Free Energy (G)

G, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it.

G < 0 for a spontaneous processG > 0 for a nonspontaneous processG = 0 for a process at equilibrium

STANDARD FREE ENERGY OF FORMATIONG°f

The free energy change that occurs when 1 mol of substance is formed from the elements in their standard state.

Calculate G° for:

2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)

Table of Gf

INTERPRETING G° FOR SPONTANEITY

1. When G° is very small (less than -10 KJ) the reaction is spontaneous as written. Products dominate.

G° < 0 G°(R) > G°(P)

2. When G° is very large (greater than 10 KJ) the reaction is non spontaneous as written. Reactants dominate.

G° > 0 G°(R) < G°(P)

3. When G° is small (+ or -) at equilibrium then both

reactants and products are present.G° = 0

Ba(OH2)•8 H2O(g) + 2 NH4NO3(g) 2 NH3(g)+10 H2O(l) + Ba(NO3)3(aq)

27

Free Energy Change and Spontaneity

Ex. 17.3a – The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has H = +95.7 kJ and S = +142.2 J/K at 25°C.

Calculate G and determine if it is spontaneous.

Since G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.

H = +95.7 kJ, S = 142.2 J/K, T = 298 K

G, kJ

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

GT, H, S

STHG

J 1033.5

2.142K 298J 1095.7

STHG

4

K

J3

Ex. 17.3a – The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has H = +95.7 kJ and S = +142.2 J/K.

Calculate the minimum temperature it will be spontaneous.

The temperature must be higher than 673K for the reaction to be spontaneous

H = +95.7 kJ, S = 142.2 J/K, G < 0

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

TG, H, S

STHG

K

J3

K

J3

2.142TJ 1095.7

02.142TJ 1095.7

0 STHG

TK 673

T 2.142

J 1095.7

K

J

3

GIBBS FREE ENERGY : GG = H - TS

describes the temperature describes the temperature dependence of spontaneitydependence of spontaneity

Standard conditions (1 atm, if soln=1M & 25°):G° = H° - TS°

A process ( at constant P & T) is spontaneous in the direction in which the free energy decreases.

1. Calculate H°, S° & G° for 2 SO2(g) + O2(g) 2 SO3(g) at 25°C & 1 atm

31

Example - G• Calculate G at 427°C for the reaction below if the

PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0 atm

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

G AND EQUILIBRIUM

The equilibrium point occurs at the lowest free energy available to the reaction system.

When a substance undergoes a chemical reaction, the reaction proceeds to give the minimum free energy at equilibrium.

G = G° + RT ln (Q)at equilibrium: G = 0

G° = -RT ln (K) G° = 0 then K = 1 G° < 0 then K > 1 G° > 0 then K < 1

Q: Corrosion of iron by oxygen is 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

calculate K for this Rx at 25°C.

G = G° + RT ln (Q)

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Free Energy, Equilibrium and Reaction Direction

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

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

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

G = RT ln Q/K = RT lnQ - RT lnK

Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and ln Q = 0 so

G0 = - RT lnK

FO

RW

AR

D R

EA

CT

ION

RE

VE

RS

E R

EA

CT

ION

Table 1 The Relationship Between G0 and K at 250C

G0(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

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Example - K• Estimate the equilibrium constant and position of

equilibrium for the following reaction at 427°C

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

Practice Problems on Go & K

1.Calculate Gº at 25ºCAgI (s) Ag+ (aq) + I- (aq)

BaSO4 (s) Ba2+(aq) + SO4

2-(aq)

What is the value for Ksp at 25ºC?

2.Calculate K at 25ºC for Zn(s) + 2H+

(aq) Zn2+(aq) + H2 (g).

Table

38

Temperature Dependence of K• for an exothermic reaction, increasing the

temperature decreases the value of the equilibrium constant

• for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant

R

S

T

1

R

Hln rxnrxn

K

Gº & Spontaneity is dependent on Temperature

Hº Sº Gº

- + - Spontaneous at all T

+ - + Non spontaneous at all T

- - +/- At Low T= Spontaneous

At High T= Nonspontaneous

+ + +/- At low T= Nonspontaneous

At High T= Spontaneous

Q. Predict the Spontaneity for H2O(s) H2O(l)

at (a) -10ºC , (b) 0ºC & (c) 10ºC.

Practice Problems on Temperature Relationships

1. At what temperature is the following process spontaneous at 1 atm?

Br2 (l) Br2 (g)

What is the normal boiling point for Br2 (l)?

2. Calculate Gº & Kp at 35ºC

N2O4 (g) 2 NO2 (g)

3. Calculate Hº, Sº & Gº at 25ºC and 650ºC.

CS2 (g) + 4H2 (g) CH4 (g) + 2H2S(g)

Compare the two values and briefly discuss the spontaneity of the Rx at both temperature.