Entropy and Free Energy - Suffolk County Community · PDF filePage 1 Jeffrey Mack California State University, Sacramento Chapter 19 Principles of Chemical Reactivity: Entropy and

Page 1

Jeffrey Mack

California State University,

Sacramento

Chapter 19 Principles of Chemical Reactivity: Entropy and Free Energy

Chemical Kinetics provides us with information about the rate of reaction and how the reaction proceeds.

Chemical Thermodynamics provides us with information about equilibrium and whether or not a reaction is Spontaneous.

Entropy & Free Energy

Chemical Thermodynamics provides us with information

about equilibrium and whether or not a reaction is

Spontaneous.

Spontaneous changes occur only in the direction

that leads to equilibrium.

Systems never change spontaneously in a direction

that takes them farther from equilibrium.

Example: Heat Transfer

Heat (thermal energy) always flows spontaneously from a

hot object to a cold object.

The process occurs until thermal equilibrium is achieved,

that is when both objects are at the same temperature.

Entropy & Free Energy

Additional Examples:

• Solvation of a Soluble salt: NH4NO3(s) dissolves

spontaneously even the process is endothermic (H

> 0)

• Expansion of a gas and Diffusion. Gasses mix to

form homogeneous mixtures spontaneously.

• Certain Phase Changes: Ice melts spontaneously

above 0oC. Water evaporates even though the

enthalpy of vaporization is endothermic.

• Certain Chemical Reactions: Na(s) reacts

vigorously when dropped in water. Iron rusts when

exposed to the atmosphere.

Spontaneous Process

But many spontaneous reactions or processes are endothermic or even have ∆H 0.

NH4NO3(s) + heat NH4NO3(aq)

∆H = 0

Spontaneous Process

A Review of Concepts of Thermodynamics

• First law of thermodynamics: The law of conservation

of energy; energy cannot be created or destroyed.

• State Function: Quantity in which its determination is

path independent.

• U = q + w: The change in internal energy of a system

is a function of heat and work done on or by the system.

• H: Heat transferred at constant pressure.

• Exothermic Process: H < 0

• Endothermic Process: H > 0

Thermodynamics

Page 2

Enthalpy alone does not predict spontaneity:

Some processes are energetically favored (rH < 0)

but not spontaneous.

Equilibrium alone cannot determine spontaneity:

Some processes are favored based on Equilibrium (K

>> 1) yet they are non-spontaneous.

There must be another factor that plays a role in

determination of spontaneity!

Thermodynamics

Diamond is thermodynamically favored to convert to graphite, but not kinetically favored.

Paper burns once the

reaction is initiates. The

process is product-favored

& kinetically favored.

Thermodynamics & Kinetics

Reactants

Products

Kin

etic

s

Thermodynamics


Factors that Affect Spontaneity

(Thermodynamic favorability):

1. Enthalpy: Comparison of bond energy (H)

2. Entropy: Randomness vs. Order of a system

(S)

In general, enthalpy is more important

than entropy.


• In a spontaneous processes the

energy of the final state is more

dispersed.

• The system moves to a higher state

of disorder.

• The thermodynamic quantity

associated with disorder and

energy dispersal is called

ENTROPY, S.

• The 2nd law of thermodynamics

states that a spontaneous process

results in an increase in the entropy

of the universe. S > 0

Reaction of K

and water

Dispersal of Energy: Entropy

Observation of a spontaneous process shows that it is associated with a dispersal of energy.

Energy Dispersal


Page 3

The change in entropy for a spontaneous process is

given by:

Where qrev is the heat gained or lost by the system

during the process and T is the absolute temperature.

A reversible process can be returned to its original

state. (chemical equilibrium), an irreversible process

cannot.

Example: The breaking of a coffee mug into many

pieces is an irreversible process.

revqS

TD =


To begin, particle 1 has 2 units of energy and 2-4

have none.


Particle 1 can transfer one unit of energy to particle 2,

then the other to 3 or 4.


Particle 2 can initially have two units of energy.


Particle 2 transfer one unit to particles 4 or 3. (2 to 1

has already been counted.)


Particle 4 can initially have two units of energy.


Page 4

Particle 4 transfer one unit to particle 3. (4 to 1 & 4 to

2 have already been counted.)


Particle 3 can start with two units of energy. Energy

transfers between particle 3 were previously counted.


Each unique combination that results in a dispersion of energy

is called a microstate. There are 10 microstates in this

system. The greater the number of microstates, the greater the

entropy of the system.

Dispersal of Energy: Entropy Dispersal of Energy: Entropy

As the size of the

container increases, the

number of microstates

accessible to the system

increases. Therefore the

entropy of the system

increases.


• The entropy of liquid

water is greater than

the entropy of solid

water (ice) at 0˚ C.

• Energy is more

dispersed in liquid

water than in solid

water due to the lack

of an ordered network

as in the solid state.


Page 5

So (J/K•mol)

H2O(liq) 69.95

H2O(gas) 188.8

Energy dispersal

S (solids) < S (liquids) < S (gases)

Entropy & States of Matter

S˚(Br2 liq) < S˚(Br2 gas) S˚(H2O sol) < S˚(H2O liq)

Entropy & States of Matter

Entropy and Microstates:

As the number of microstates increases, so

does the entropy of the system.

S = klnW

k = Boltzman’s constant (1.381 1023 J/K)

W = the number of microstates


Entropy and Microstates:

The change in entropy associated with a

process is a function of the number of final and

initial microstates of the system.

If Wfinal > Winital, S > 0

If Wfinal < Winital, S < 0

final initial

final final

final

initial

S S S

S k lnW k lnW

WS k ln

W

D = =

D = × - ×

æ öD = × ç ÷

è ø

Entropy, Entropy Change, & Energy Dispersal: A Summary

When a solute dissolves in a solvent the process is

spontaneous owing to the increase in entropy.

Matter (and energy) are more dispersed. The number

of microstates is increased.


The entropy of a substance increases with temperature.

Molecular motions of

heptane, C7H16

Molecular motions of

heptane at different temps.

Entropy Measurements & Values

Page 6

An Increase in molecular complexity generally leads to increase in S.


Entropies of ionic solids depend on coulombic attractions.

S° (J/K•mol)

MgO 26.9

NaF 51.5

Mg2+ & O2- Na+ & F-


Defined by Ludwig Boltzmann, the third law states that a

perfect crystal at 0 K has zero entropy; that is, S =0.

The entropy of an element or compound under any other

temperature and pressure is the entropy gained by converting

the substance from 0 K to those conditions.

To determine the value of S, it is necessary to measure the

energy transferred as heat under reversible conditions for the

conversion from 0 K to the defined conditions and then to use

Equation 19.1

Because it is necessary to add energy as heat to raise the

temperature, all substances have positive entropy values at

temperatures above 0 K.

revqS

TD =

Standard Molar Entropies

The standard molar entropy, S°, of a substance is the entropy

gained by converting 1 mol of it from a perfect crystal at 0 K to

standard state conditions (1 bar, 1m for a solution) at the specified

temperature.

Standard Molar Entropies

rev

boil

q 40,700 J/mol JS 109

T 373.15 K mol KD = = = +

×

For a phase change,

where q = heat transferred in the

phase change

revqS

TD =

2 2For H O (liq) H O(g)

JH q 40,700

molD = = +

Entropy Changes for Phase Changes

S increases

slightly with T

S increases a large

amount with phase

changes

Entropy & Temperature

Page 7

Standard molar entropy values can be used to

calculate the change in entropy that occurs in various

processes under standard conditions.

The standard entropy change for a reaction (rS°)

can be found in the same manner as rH° were:

Where n & m are the stoichiometric balancing

coefficients.

This calculation is valid only under reversible

conditions.

S n S (products) m S (reactants)

Determining Entropy Changes in Physical & Chemical Processes

Problem: Consider the reaction of hydrogen and oxygen to form liquid

water.

What is the standard molar entropy for the reaction?



water.


rS n S (products) m S (reactants)



water.


r

2 2 2

r


2H (g) O (g) 2H O(l)

J J J JS 2 69.95 2 130.7 205.1 326.6

K K K K



water.


The enthalpy change is negative (net decrease in dispersion)

due to the change in number of moles: 3 reduced to 2

2 2 2

o

r

2H (g) O (g) 2H O(l)

J J J JS 2 69.6 2 130.7 205.3 326.9

K K K K

+ ®

æ öD = ´ - ´ + = -ç ÷

è ø


The second law of thermodynamics states:..

S°universe = ∆S°system + ∆S°surroundings

Any change in entropy for the system plus the entropy

change for the surroundings must equal the overall

change in entropy for the universe.

A process is considered to be spontaneous under

standard conditions if S°(universe) is greater than

zero.

2nd Law of Thermodynamics

Page 8

The solution process for NH4NO3 (s) in water is an entropy driven process.

∆S°universe =

∆S°system +

∆S°surroundings


Calculating S°Surroundings:



system surroundings

system surroundings

q qS S

T T



system surroundings

system surroundings

q qS S

T T

system surroundings system surroundingsq q 0 q q+ = = -



at constant pressure, qsystem = rH°system

system surroundings

system surroundings

q qS S

T T




at constant pressure, qsystem = rH°system

system surroundings

system surroundings

q qS S

T T



surroundings r system

r system

surroundings

Therefore : q H

Hand S

T

Page 9

Problem:

Calculate the entropy change for the surroundings for

the reaction: 2H2(g) + O2(g) 2H2O(l)


Problem:



The enthalpy change for the reaction is calculated

using standard molar enthalpies of formation:


r system

surroundings

HS

T

Problem:



The enthalpy change for the reaction is calculated

using standard molar enthalpies of formation:


r system

surroundings

HS

T

3

surroundings

10 K571.7kJ

1kJ JS 1917K298.15K

Problem:



As calculated previously:

S°universe = 327 J/K + 1917 J/K =1590. J/K

surroundingsJS 1917K

systemJS 327K


Problem:



S°universe = 327 J/K + 1917 J/K = 1590. J/K

The entropy of the universe increases ( > 0) therefore

the process is spontaneous at standard state

conditions.

The process is spontaneous due to the entropy

change in the surroundings, not the system.

2nd Law of Thermodynamics Spontaneous or Not?

Page 10

The method used so far to determine whether a

process is spontaneous required evaluation of two

quantities, S°(system) and S°(surroundings).

J. Willard Gibbs asserted that that the

maximum non-PV work available to the

system must be a function of Enthalpy

and Entropy:

G = H – TS

G = Gibbs Free energy of the system, H = system

enthalpy and S = the entropy of the system.

J. Willard Gibbs

1839-1903

Gibbs Free Energy, G

Since it is impossible to measure individual values of

enthalpy, we often express free energy in terms of the

changes of thermodynamic quantities.

(G = H – TS)

At constant temperature:

G° = Gibbs Free energy change of the system,

H° = enthalpy change and S° = the entropy

change at SS conditions.

G H T S


If the reaction is exothermic (H < 0)

And the change in entropy is positive (H > 0)

at a given temperature, then S < 0.

We then can assert that if G < 0 that the

reaction is spontaneous as well as product

favored!

Spontaneity is a function of energy and

dispersion!

G H T S


H S G Reaction

+ Product Favored

+ + Reactant Favored

? Temperature dependant

+ + ? Temperature dependant


G H T S

Since rG° is related to S°universe, it follows

that:

• If rG° < 0: The process is spontaneous in the

direction written under standard conditions.

• If rG° = 0: The process is at equilibrium under

standard conditions.

• If rG° > 0: The process is non-spontaneous in

the direction written under standard conditions.

• Conclusion: A reaction proceeds spontaneously

toward the minimum in free energy, which

corresponds to equilibrium.

rGo & Equilibrium

Product Favored Reactions, ∆G° negative, K > 1

Q < K: Heading to

equilibrium G < 0

Q = K: At

equilibrium G = 0

Q > K: Heading

away from

equilibrium G > 0

∆G, ∆G°, Q, & K

Page 11

• Product-favored

• 2 NO2 N2O4

• ∆rG° = – 4.8 kJ

• State with both reactants

and products present is

more stable than

complete conversion.

• K > 1, more products

than reactants.

∆G, ∆G°, Q, & K

Reactant Favored Reactions, ∆G° positive, K < 1

Q < K: Heading to

equilibrium G < 0

Q = K: At

equilibrium G = 0

Q > K: Heading

away from

equilibrium G > 0

∆G, ∆G°, Q, & K

• Reactant-favored

• N2O4 2 NO2

∆rG° = +4.8 kJ

• State with both

reactants and products

present is more stable

than complete

conversion.

• K < 1, more reactants

than products

∆G, ∆G°, Q, & K

rG° represents the free energy change for a

process at standard state conditions. (Equilibrium)

What if this is not the case?

Under nonstandard conditions:

Where R is the

gas law constant

and Q =

r rG G RTln(Q)

[ ] [ ]

[ ] [ ]

c d

a b

C DQ

A B

for aA + bB cC dD

=

+

∆G, ∆G°, Q, & K

At equilibrium, we know that rG = 0 and Q = K

Therefore:

So knowing one quantity yields the other.

r r

r

r

G G RTln(Q)

0 G RTln(K)

G RTln(K)

∆G, ∆G°, Q, & K

Problem:

rG° for the formation of ammonia at SS conditions is 16.37

kJ/mol. What is the value of Kp at this temperature and

pressure? (1 mol, 20 °C and 1 atm)

∆G, ∆G°, Q, & K

Page 12

Problem:




2 2 3

r p

3 1H (g) N (g) NH (g)

2 2

G RTln(K )

∆G, ∆G°, Q, & K

Problem:




r p

r

3

rp

G RTln(K )

Gln(K)

RT

kJ 10 J16.37

G mol 1 kJK exp exp 740.JRT 8.314 298 K

mol K

∆G, ∆G°, Q, & K

Problem:




r p

r

3

rp

G RTln(K )

Gln(K)

RT

kJ 10 J16.37

G mol 1 kJK exp exp 740.JRT 8.314 298 K

mol K

K >> 1, product

favored as

predicted by G

∆G, ∆G°, Q, & K

The relation of ∆rG, ∆rG°, Q, K, reaction spontaneity,

and product- or reactant favorability.

Summary

C (graphite) + 2H2(g) CH4(g)

rH° (kJ/mol)

0 0 74.9

S° (J/K) +56 +130.7 + 186.3

Calculating & Using Free Energy Standard Free Energy of Formation

r f f

r

H n H (products) m H (reactants)



rH° (kJ/mol)

0 0 74.9

S° (J/K) +56 +130.7 + 186.3


Page 13

r r

JH 74.9 kJ S 80.7

K



rHo

(kJ/mol) 0 0 74.9

So (J/K) +56 +130.7 + 186.3

r r

r r r

r 3

r

JH 74.9 kJ S 80.7

K

G H T S

J 1 kJG 74.9 kJ 298 K 80.7

K 10 J

kJG 50.9 mol

rG° is negative at 298 K, so the reaction is

predicted to be spontaneous under standard

conditions at this temperature. It is also predicted to

be product-favored at equilibrium.


Under reversible conditions, both enthalpy and

entropy are state functions. It follows that the

Gibbs free energy must also be.

Therefore we can write that:

G H T S

r f fG n G (products) m G (reactants)


Note that ∆fG° for an element = 0

Free Energies of Formation

Using the fG° found in appendix L of your text,

calculate rG° for the following reaction:

4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)




4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)



Page 14



4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)



r f 2 f 2 f 2G 4 G NO (g) 6 G H O(g) 4 G NO (g) 0



4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)

rG° = –1101.14 kJ/mol (product-favored

r f 2 f 2 f 2

r

G 4 G NO (g) 6 G H O(g) 4 G NO (g) 0

G 4 mol (51.23 kJ/mol) 6 mol ( 228.59 kJ/mol)

4 mol ( 16.37 kJ /mol)



Is the dissolution of ammonium nitrate product-

favored?

If so, is it enthalpy- or entropy-driven?

NH4NO3(s) + heat NH4NO3(aq)

Calculating ∆rG°

• By definition:

G = H − TS

• Indicating that free energy is a function of temperature.

• rG° will therefore change with temperature.

• A consequence of this temperature dependence is that, in certain instances, reactions can be product-favored at equilibrium at one temperature and reactant-favored at another.

Free Energy & Temperature

When a reaction has

rH° < 0

&

rS° > 0

at all temperatures

rG° is negative.

(Product favored)


When a reaction has

rH° > 0

&

rS° > 0

at high temperatures

rG° is negative.

(Product favored)

When a reaction has

rH° < 0

&

rS° < 0

at low temperatures

rG° is negative.

(Product favored)


Page 15


At what temperature will a reaction that is non-spontaneous

turn over to a reaction that is? (rG changes from + to )




Example: The reaction,

Has the following thermodynamic values.

∆rH° = +470.5 kJ

∆rS° = +560.3 J/K

∆rG° = +301.3 kJ

Reaction is reactant-favored at 298 K

2 3 22Fe O (s) 3C(s) 4Fe(s) 3CO (g)


2 3 22Fe O (s) 3C(s) 4Fe(s) 3CO (g)+ +





When rG 0, the reaction begins to become spontaneous.

2 3 22Fe O (s) 3C(s) 4Fe(s) 3CO (g)

2 3 22Fe O (s) 3C(s) 4Fe(s) 3CO (g)+ +




When rG 0, the reaction begins to become spontaneous.

2 3 22Fe O (s) 3C(s) 4Fe(s) 3CO (g)

r r r

r

r

G H T S 0

HT

S





∆rH° = +467.9 kJ

∆rS° = +560.3 J/K

3

r

r

10 J467.9 kJ

H 1 kJT 839.7 KJS

560.3K


2 3 22Fe O (s) 3C(s) 4Fe(s) 3CO (g)

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Entropy and Free Energy - Suffolk County Community · PDF filePage 1 Jeffrey Mack California State University, Sacramento Chapter 19 Principles of Chemical Reactivity: Entropy and