Chemical Thermodynamics Chapter 17 Chemical Thermodynamics.

ChemicalThermodynamics

Chapter 17 Chemical Thermodynamics


Spontaneous Processes

Entropy

Second Law of Thermodynamics

Third Law of Thermodynamics

Gibbs Free Energy

Predicting Spontaneity

Standard Enthalpies of Formation

Gibbs Free Energies of Formation

Free Energy Changes

Contents


• Thermodynamics is

the study of energy

relationships that

involve heat,

mechanical work, and

other aspects of

energy and heat

transfer.


First Law of Thermodynamics• You will recall from Chapter 6 that energy

cannot be created nor destroyed.

• Therefore, the total energy of the universe is a constant.

• Energy can, however, be converted from one form to another or transferred from a system to the surroundings or vice versa.


First law of thermodynamics, the law of

conservation of energy, dictates the relationship

between heat (q), work (w), and changes in internal

energy ( ΔU ).

ΔU = q + w

Notice: assigning the correct signs to the quantities of heat and work.


Spontaneous Processes• Spontaneous processes

are those that can

proceed without any

outside intervention.

• The gas in vessel B will

spontaneously effuse into

vessel A, but once the

gas is in both vessels, it

will not spontaneously

separate from each other.


Spontaneous Processes

Processes that are

spontaneous in one

direction are

nonspontaneous in

the reverse

direction.


Spontaneous Processes• Processes that are spontaneous at one

temperature may be nonspontaneous at other temperatures.

• Above 0C it is spontaneous for ice to melt.• Below 0C the reverse process is spontaneous.


Goal of chemical thermodynamics:

predicting which changes will be

spontaneous.


Reversible Processes

In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.


Irreversible Processes

• Irreversible processes cannot be undone by

exactly reversing the change to the system.

• Spontaneous processes are irreversible.


Entropy

• Entropy (S) is a term coined by Rudolph Clausius in

the 19th century.

• Clausius was convinced of the significance of the

ratio of heat delivered and the temperature at which

it is delivered, qT


Entropy

• Entropy can be thought of as a measure of the

randomness of a system.

• It is related to the various modes of motion in

molecules.

• The entropy of a system in a given state is a

measure of the number of different microscopic

states that correspond to a given macroscopic

state.


The more the particles and their positions, the more disordered the system is.


Entropy on the Molecular Scale

• The number of microstates and, therefore, the entropy tends to increase with increases inTemperature.Volume.The number of independently moving

molecules.


Entropy

• Like total energy, U, and enthalpy, H,

entropy is a state function.

• Therefore,

S = Sfinal Sinitial


Entropy

• For a process occurring at constant temperature (an

isothermal process), the change in entropy is equal

to the heat that would be transferred if the process

were reversible divided by the temperature:

S =qrev

T


Entropy and Physical States

• Entropy increases with

the freedom of motion

of molecules.

• Therefore,

S(g) > S(l) > S(s)



The second law of thermodynamics states

that the entropy of the universe increases for

spontaneous processes, and the entropy of

the universe does not change for reversible

processes.


S and Isolated Systems

• For an equilibrium process in an isolated system, S = 0

• For a spontaneous process in an isolated system, S > 0



In other words:

For reversible processes:

Suniv = Ssystem + Ssurroundings = 0

For irreversible processes:

Suniv = Ssystem + Ssurroundings > 0



These last truths mean that as a result

of all spontaneous processes the

entropy of the universe increases.


Entropy Changes

• In general, entropy increases whenGases are formed from

liquids and solids.Liquids or solutions are

formed from solids.The number of gas

molecules increases.The number of moles

increases.


Solutions

Generally, when a solid is dissolved in a solvent, entropy increases.


Entropy and Disorder

• The entropy S of a system is a measure of the disorder or randomness in the system.

• Disorder can be defined as the number of equivalent ways of distributing the conserved matter and energy through the system.

• Example: a substance has higher entropy in the gaseous state than in the solid state.


Entropy Increasing Processes

• The entropy is expected to increase for

processes in which

1. liquids or solutions are formed from solids.

2. gases are formed from either liquids or solids.

3. the number of molecules of gas increases in

going from reactants to products.

4. The number of degrees of freedom increases.



The entropy of a pure crystalline

substance at absolute zero is 0.



• The entropy of all pure crystalline substances approach zero as the temperature approaches absolute zero – since all disorder has been removed.

as T 0 K, S0 0

• This defines the absolute entropy scale S0.


Standard Entropies

• The standard entropy of a substance S0 is the

entropy change required to heat 1 mole of the

substance from 0 K to the temperature of 298 K.

• Standard molar entropies are used to calculate

S for reactions, just as Hf values are used to

calculate H for reactions.

• Note that S0 for an element in its standard state

is not zero – unlike the case for Hf


Standard Entropies

• These are molar entropy

values of substances in

their standard states.

• Standard entropies tend

to increase with

increasing molar mass.


Standard Entropies

Larger and more complex molecules have greater entropies.


Entropy Changes

Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated:

Srxn° = npS°(products) - nrS°(reactants)

where n and m are the coefficients in the balanced chemical equation.


Sample problem to calculate Srxn°:

Use standard entropies to calculate Srxn°for the

reaction: 2SO2(g)+O2 (g)= 2SO3(g)

Solution:

Equation: 2SO2(g)+O2 (g)= 2SO3(g)

S°, J/K/mol 248.12 205.03 256.72


=2× S°(SO3) - 2× S°(SO2) - × S°(O2)

=2 ×256.72-2 ×248.13-205.03

=-187.83 J/K/mol


Entropy Exercise

Predict whether the entropy change in each of the

following reactions is positive or negative:

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

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

c) H2O(l) H2O(g)

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

e) 4Fe(s) + 3O2(g) 2Fe2O3(s)


Entropy Changes in Surroundings

• Heat that flows into or out of the system changes the entropy of the surroundings.

• For an isothermal process:

Ssurr =qsys

T

• At constant pressure, qsys is simply H for the system.


Entropy Change in the Universe

• The universe is composed of the system and the surroundings.

• Therefore,

Suniverse = Ssystem + Ssurroundings

• For spontaneous processes

Suniverse > 0


Entropy Change in the Universe

• This becomes:

Suniverse = Ssystem +

Multiplying both sides by T,

TSuniverse = Hsystem TSsystem

Hsystem

T


Gibbs Free Energy

TSuniverse is defined as the Gibbs free

energy, G.

• When Suniverse is positive, G is negative.

• Therefore, when G is negative, a process

is spontaneous.


Criterion for Spontaneity

• According to the 2nd Law, a reaction is

spontaneous at constant pressure and

temperature if and only if:

Hsystem TSsystem<0

• Or …

H - T S < 0


Gibbs Free Energy

These two factors are combined in the

Gibbs free energy, defined: G = H –TS or G = H - T S

A reaction is spontaneous (under conditions of

constant T and P) when G < 0

• Spontaneous reactions are favored by: H < 0 (exothermic)

S > 0 (increasing entropy)



1. If a reaction has G < 0 it is

spontaneous (in the forward direction).

2. If a reaction has G > 0 its reverse is

spontaneous.

3. If a reaction has G =0 then it is already

at equilibrium.


Gibbs Free Energies of Formation

• The Gibbs free energy of formation Gf for a

substance is defined in the same way as the

enthalpy of formation (Hf)..

Gf for a substance is the Gibbs free energy

change when one mole of the substance is formed

under standard conditions from its elements in

their standard states.


Standard Free Energy Changes

Analogous to standard enthalpies of formation are standard free energies of formation, G.

f

Grxn = npGf(products) nrGf(reactants)

where np and nr are the stoichiometric coefficients.


Grxn from Gf Values

• Just as S of reaction can be calculated from standard entropy values S0, and H of reaction can be calculated from Hf values, so Grxn of reaction can be

calculated from Gf values.

Grxn = npGf(products) nrGf(reactants)


The difference between Grxn and Grxn

Grxn(or G ): is the free energy change

that accompanies a change from reactants

in their standard states to products in their

standard states.

Grxn(or G ): is the free energy change that

accompanies a change from reactants in

nonstandard states to products in

nonstandard states.


Example for calculation of G°from H°and S°:

Use standard heat of formation and standard entropies to calculate G°for the reaction at 25 and 1atm partial ℃pressure of each gas:

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

Solution:

(1) Calculation of H° and S°:

Equation: 3H2(g)+N2 (g)= 2NH3(g)

Hf°, kJ/mol 0 0 -46.11

S°, J/K/mol 248.12 205.03 256.72

H°=npHf°, (products) - nrHf°, (reactants)

=2 ×(-46.11) - 3×0 - 2×0

=-92.22 kJ/mol



=2× S°(NH3) - 3× S°(H2) - 2× S°(N2)

=2 ×192.3 - 3 ×130.68 - 2 × 191.50

=-198.9 J/K/mol

(2)Calculation of G°:

G°=H°-TS°= -92.22 –(273.2+25) ×(-198.9 ×10-3)

=-32.91kJ/mol


Example for G0 of Reaction from Gf0

• Calculate the standard Gibbs free energy of

reaction for the combustion of methane CH4.

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

Gf (CH4 ,g ) = -50.8 kJ/mol

Gf (CO2 ,g) = -393.4 kJ/mol

Gf (H2O ,l) = -237.13 kJ/mol

G0 = 2×Gf (H2O ,l) + Gf (CO2 ,g) - Gf (CH4 ,g )

G0 = -816.9 kJ


S ° and H ° for Reactions

where np is the stoichiometric coefficient of product,

nr is the stoichiometric coefficient of reactant.


Hrxn° = npHf°(product) - nrHf°( reactant)


Free Energy Changes

At temperatures other than 25°C,

G° = H TS

How does G change with temperature?


Free Energy and Temperature

• There are two parts to the free energy

equation:

H— the enthalpy term

TS — the entropy term

• The temperature dependence of free energy,

then comes from the entropy term.


Example for G0 from H0 and S0

• Consider N2(g) + 3H2(g) 2NH3(g). Assume that

H0 and S0 do not change much with

temperature. Calculate G0 for the reaction at

500 K.

H0 = -92.38 kJ/mol

S0 = -198.3 J/K-mol

G0 = H0 - T S0 = (-92.38) - (500) ×(-0.1983)

G0 = +6.77 kJ


• S0: depend markedly on temperature.

S0: however, because increasing temperature

increase the entropy of all substance, S0 often do

not change greatly with temperature at ordinary

temperature.

H0: are also often quite constant as temperature

changes because of the same formation and

cleavage in a certain reaction.



• Two factors determine the spontaneity

of a chemical or physical change:Enthalpy change H

Entropy change S

H < 0 (exothermic) favors the process.

S > 0 (more randomness) favors the

process.


H, S and Spontaneity

• There are four possible combinations of

positive and negative H and S:

1. H < 0 and S > 0 G < 0 : spontaneous at any

temperature

2. H > 0 and S < 0 G > 0 : not spontaneous at

any temperature

3. H > 0 and S > 0 favored at high T

4. H < 0 and S < 0 favored at low T


Temperature Dependence

• Since G=0 at equilibrium, a process will

reach equilibrium when H = T S or at

the temperature T = H/S.

• If a process is non spontaneous at T<

H/S, it will become spontaneous for T >

H/S and vice versa.


Free Energy and Temperature


Energy and EquilibriumE

nerg

y

reactants productsequilibrium

Q: reaction quotient; K: equilibrium constant

Equilibrium: A system’s macroscopic properties do

not change spontaneously. (Vforward =Vreverse)

Q < Kc means the reaction will go spontaneously in the forward direction.Q > Kc means the reaction will go spontaneously in the reverse direction.


Gibbs Free Energy

1. If G is negative, the forward reaction is spontaneous.

2. If G is 0, the system is at equilibrium.

3. If G is positive, the reaction is spontaneous in the reverse direction.


Free Energy and Equilibrium

Under any conditions, standard or

nonstandard, the free energy change can

be found this way:

G = G + RT lnQ

(Under standard conditions, all concentrations are 1

M, so Q = 1 and lnQ = 0; the last term drops out.)


Free Energy and Equilibrium

• At equilibrium, Q = K, and G = 0.

• The equation becomes

0 = G + RT lnK

• Rearranging, this becomes

G = RT lnK

or,

K = eG/RT

What's Free about Free Energy

• The change in the Gibbs free energy for a

process is the maximum amount of useful

work that can be done by the system at

constant temperature and pressure.

maxwG

CaCO3 Example

• At what temperature will CaCO3(s) just begin

to decompose to CaO(s) and CO2(g) under

standard conditions?

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

H0 = +178.3 kJ

S0 = +159.0 J/K = 0.159 kJ/K

G0 = 0 at equilibrium.

H0 - T S0 = 0 T = H0 / S0 = 1121 K

∴When T>1121K, the reaction is spontaneous.

K from G Example

Consider N2(g) + 3H2(g) 2NH3(g). Calculate

the equilibrium constant at 500 K.

H0 = -92.38 kJ/mol

S0 = -198.3 J/K-mol

G0 = H0 - T S0 = (-92.38) – (500)(-0.1983)

G0 = +6.77 kJ

K = exp(- G0 / RT) = 0.196 (K = eG/RT)

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Chemical Thermodynamics Chapter 17 Chemical Thermodynamics.

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