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Entropy and the Second Law of Thermodynamics Chapters 10

• Spontaneity

• Entropy

• The Second Law of Thermodynamics

• The Third Law of Thermodynamics

• Gibbs Free Energy

• Free Energy and Chemical Reactions

Spontaneous Process

Are the following spontaneous or nonspontaneous processes

a) The reaction between NaOH(aq) and HCl(aq)

b) The decomposition of water to give H2(g) and O2(g)

c) The melting of a ice cube

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Spontaneity: Natures Arrow

• Some processes or reactions proceed in only one direction.– Gasoline reacts spontaneously with oxygen to form carbon dioxide and

water, but water and carbon dioxide never spontaneously react to reform gasoline.

• A spontaneous process takes place without continuous intervention, according to thermodynamics.– Spontaneous processes are not necessarily rapid processes.– The combustion of diamond is thermodynamically spontaneous, but

diamonds are considered to last forever.

• Some spontaneous reactions only occur once they are initiated.– The combustion of gasoline is a spontaneous reaction but only occurs

when the reaction is initiated with a spark.

• Nonspontaneous reactions only occur with a continual input of energy.

Enthalpy and Spontaneity

• Exothermic reactions are generally preferred over endothermic reactions.

– Melting ice is an endothermic process but occurs spontaneously at room temperature.

– Enthalpy is not the exclusive determinant of spontaneity.

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Thermodynamic Example

Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected to an evacuated 4.0 L sealed bulb with a valve. Assume the temperature is constant.

a) What should happen to the gas when you open the valve?

b) Calculate H, E, q, and w for the process

c) Given your answer to part b, what is the driving force for the process?

Entropy

• Entropy is a state function and was first introduced in considering the efficiency of steam engines.– The Carnot cycle uses a

combination of adiabaticprocesses (no heat is exchanged) and isothermal processes (temperature is constant).

• For the Carnot cycle, the sum of q/T around the closed path is equal to zero and therefore defines the state function.– The state function is entropy.

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Spontaneous Processes and Entropy

• Thermodynamics lets us predict whether a process will occur but gives no information about the amount of time required for the process.

• The driving force for a spontaneous process is an increase in the entropy of the universe.

• Entropy, S, can be viewed as a measure of randomness, or disorder.

Definition of Entropy

• Entropy can be tentatively defined as a measurement of the randomness, or disorder, of a system.– For large numbers of particles, probability favors random

arrangements.– Statistical mechanics or statistical thermodynamics provides a

quantitative basis and molecular perspective to entropy using probability.

• For entropy, the probability for the number of ways in which particles can achieve the same energy is used.– The way by which a collection of particles can assume a given

energy is a microstate.– The number of possible microstates is designated by .– The number of microstates increases as the “randomness” of the

system increases.

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Defining Entropy

• The Maxwell-Boltzmann distribution indicates the overall collection of molecular speeds but not the speed of individual particles.

• Energy is exchanged during molecular collisions, without disrupting the overall distribution of speeds.

The Boltzmann Equation for Entropy

• As the number of microstates for a system increases, the entropy of the system increases. This relationship is defined by the equation:

S = kb ln

– S is entropy

– kb is the Boltzmann constant.• The ideal gas constant per molecule = R/NA

– is the number of microstates.• The particular way in which particles are distributed amongst

the states. Number of microstates =

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Probability and Spontaneous Change

• Probability has important uses in chemistry.

• Consider 2 connected bulbs containing two molecules.

• Each arrangement (microstate) has an equally probability of occurring– 25% chance of both molecules being

in the left bulb

– 50% chance of one molecules being in the each bulb

– 25% chance of both molecules being in the right bulb

Possible arrangements (states) of four molecules in a two-bulbed flask.

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Chance of an Arrangement in Two Flask System

0%

5%

10%

15%

20%

25%

30%

35%

40%

0% 20% 40% 60% 80% 100%% of Molecules in Left Flask

% O

cc

urr

en

ce

N=4

N=10

N=100

N=1000

Judging Entropy Changes in Processes

• Certain types of changes will result in increased entropy.

– Certain phase changes.

• Ssolid < Sliquid << Sgas

– An increase in the number of particles present.

– An increase in the temperature of a substance.

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Judging Entropy Changes in Processes

• When a solid melts to form a liquid, entropy increases.– In solids, the particles are held in place rigidly, limiting the number

of ways a specific energy can be obtained.– In liquids, the particles move past each other, increasing the

number of ways a specific energy can be obtained.– The number of microstates increases during melting.

• A chemical reaction that generates two moles of gas when only one mole of gas was initially present will increase the entropy of a sample.– The number of possible microstates increases as the number of

particles increases.

• When a sample is heated, the temperature of the sample increases.– As temperature increases, the number of possible velocities

increases.

Entropy Change

ΔS = qrev

T

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Entropy Changes

• The greater the degree of randomness or disorder in a system, the greater the entropy of the system

• Entropy increases are expected to accompany processes in which– Pure liquids or liquids solutions are formed from solids

– Gases are formed, either from solids or liquids

– The number of molecules of gases increase during the reaction

– The temperature of a substance is increased (increased temperature means increased molecular motion)

Entropy Change Examples

Predict whether each of the following processes involves an increase, decrease or uncertain change in entropy

1. 2NH4NO3(s) 2N2(g) + 4H2O(g) + O2(g)

2. 2SO2(g) + O2(g) 2SO3(g)

3. NaCl(aq) NaCl(s)

4. CO(g) + H2O(g) CO2(g) + H2(g)

5. CO(g) + H2O(l) CO2(g) + H2(g)

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Criteria for Spontaneous Change:The Second Law of Thermodynamics.

ΔStotal = ΔSuniverse = ΔSsystem + ΔSsurroundings

The Second Law of Thermodynamics:

ΔSuniverse = ΔSsystem + ΔSsurroundings > 0

All spontaneous processes produce an increase in the entropy of the universe.

Criteria for Spontaneous Change:The Second Law of Thermodynamics.

• The second law of thermodynamics: in any spontaneous process, the total entropy of the universe is positive– (Su > 0).

Su = Ssys + Ssurr > 0• Su = entropy of the universe

• Ssys = entropy of the system

• Ssurr = entropy of the surroundings

– It is impossible to convert heat completely to work, since work is a process that involves moving random motions into more ordered ones.

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Implications and Applications

• The entropy change for the surroundings can be calculated from the heat flow from the system, which is equal to –H.

– For an exothermic reaction, the entropy of the surroundings increases.

Ssurr H

T

The Third Law of Thermodynamics

• The third law of thermodynamics states that the entropy of a perfect crystal of any pure substance approaches zero as the temperature approaches absolute zero.

• Because S is explicitly known (= 0) at 0 K, S values at other temps can be calculated.

• The entropy of one mole of a chemical substance under standard conditions is the standard molar entropy, So.

• The entropy change for a reaction can be calculated from the standard molar entropies of the reactants and products.

j

jji

ii ntsreactaSproductsSS

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(a) A perfect crystal of hydrogen chloride at 0 K.

(b) As the temperature rises above 0 K, lattice vibrations allow some dipoles to change their orientations, producing some disorder and an increase in entropy.

The H2O molecule can vibrate and rotate in several ways, some of which are shown here.

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Entropy as a Function of Temperature

The Third Law of Thermodynamics

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Implication of the Third Law of Thermodynamics

• Entropy is a state function, so the value of S must be independent of the path taken from reactants to products.

• The entropy change for a reaction can be calculated from the standard molar entropies of the reactants and products.

j

jji

ii ntsreactaSproductsSS

S° Examples

• What is S° for the reaction

2NO(g) + O2(g) 2NO2(g)

Given: S°NO = 211 J mol-1 K-1

S°O₂ = 205 J mol-1 K-1

S°NO₂ = 240 J mol-1 K-1

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Free Energy and Free Energy Change

• Hypothetical process:

– only pressure-volume work, at constant T and P.

• Make the enthalpy change reversible.

– large surroundings, infinitesimal change in temperature.

• Under these conditions we can calculate entropy.

qsur = -qp = -ΔHsys

syssursyssur

sur HSTorT

H

T

qS

Free Energy and Free Energy Change

TΔSu. = TΔSsys + TΔSsur

TΔSu. = TΔSsys – ΔHsys = -(ΔHsys – TΔSsys)

-TΔSu. = ΔHsys – TΔSsys

G = H – TS (Gibbs Free Energy)

ΔG = ΔH – TΔS (at constant T)

For the universe:

For the system:

ΔGsys = - TΔSu

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Criteria for Spontaneous Change at Constant Temperature and Pressure

ΔGsys < 0 (negative), the process is spontaneous.

ΔGsys = 0 (zero), the process is at equilibrium.

ΔGsys > 0 (positive), the process is non-spontaneous.

Free Energy and Spontaneous Change

• For the last two cases, the temperature at which a reaction changes from

spontaneous to nonspontaneous can be calculated.

S

HT

STHG

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Free Energy and Spontaneous Change

• For a spontaneous reaction with a negative H and a negative S, the reaction only occurs at low temperature. – These reactions are referred to as enthalpy driven because the

negative value for the enthalpy is responsible for the negative value of the Gibbs free energy change.

• For a spontaneous reaction with a positive H and a positive S, the reaction only occurs at high temperature.– These reactions are referred to as entropy driven because the

product of the positive entropy change and the absolute temperature is responsible for the negative value of the Gibbs free energy change.

Spontaneous Change Predictions

Using the enthalpy and entropy changes to predict under what conditions would the following reactions occur spontaneously

1. 2NH4NO3(s) 2N2(g) + 4H2O(g) + O2(g) H = -236kJ/mol

2. I2(g) 2I(g)

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Melting of Ice

Free Energy and Work

• It can be shown that the Gibbs free energy change is equal to the maximum useful work done by the system.

G = -wmax

– Work is not a state function.

– Maximum work realized only if the reaction or process is carried out along a very specific path.

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Free Energy and Work

• A reversible path is the specific path required for maximum useful work.

– For systems with a reversible path, the system is near equilibrium and a small incremental change in a variable will bring the system back to its initial state.

• In an irreversible change, a small incremental change in any variable does not restore the initial state.

– The amount of work available is always less than maximum for an irreversible change.

Free Energy and Work

• The Gibbs free energy change establishes the upper bound to the amount of work obtained from a given process.

– Actual work produced in any real application may be considerably less.

• Reactant mixtures are generally very far from equilibrium.

– Systems that are far from equilibrium often change rapidly, and rapid changes tend to be irreversible.

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Standard Free Energy Change, ΔG°

• The standard free energy of formation, ΔGf°.– The free energy change for a reaction in which one mole of a

substance in its standard state is formed from its elements in their standard states.

– Gf° = 0 for elements in their free standard state.

• The standard Gibbs free energy change for a reaction, G°,can be calculated from Gibbs free energies of formation, Gf°.

– This equation provides an alternative method to calculate G°, without entropy or enthalpy.

G° = H° - TS°

j

jfji

ifi ntsreactaGproductsGG

Free Energy and Chemical Reactions

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Characteristics of G

G is an extensive property

G changes sign when a process is reversed

G for a net or overall process can be obtained by summing the G values for the individual steps

4. For a chemical reaction at constant T and P

G = H - TS

5. For a chemical reaction

ΔG° = [ npΔGf°(products) - nrΔGf°(reactants)]

G° Examples

Given that at 400°C the reaction 4Cu(s) + O2(g) 2Cu2O(s) G° = -250 kJ/mol

1. What is G° at 400°C for the following reactionsa) 2Cu(s) + ½O2(g) Cu2O(s)b) Cu2O(s) 2Cu(s) + ½O2(g)

2. Will Cu2O(s) spontaneously decompose at 400°C with the compounds in their standard states?

3. What happens if the Cu2O(s) decomposition reaction is made to run simultaneously with

C(s) + ½O2(g) CO(g) G° = -175 kJ/mol

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Coupled Reactions

• In order to drive a non-spontaneous reactions we changed the conditions (i.e. temperature or electrolysis)

• Another method is to couple two reactions.– One with a positive ΔG and one with a negative ΔG.

– Overall spontaneous process.

G° Examples

Consider the reaction 2NO(g) + O2(g) 2NO2(g)

4. Given at 25°C H° = -114.1 kJ/mol and S° = -146.3 J mol-1 K-1

What is G°?

5. Given the Gf° of NO(g) is 86.57 kJ/mol and Gf° of NO2(g) is 51.30kJ/mol What is G°?

6. Given for water at 25°C Hvap° = 42 kJ/mol and Svap° = 119 J mol-1 K-1

What is Gvap° at 25°C? Will water spontaneously evaporate at 25°C?

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Free Energy Change and Equilibrium

Relationship of ΔG° to ΔG for Nonstandard Conditions

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

ΔG = ΔH - TΔS ΔG° = ΔH° - TΔS°

For ideal gases ΔH = ΔH°

ΔG = ΔH° - TΔS

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Relationship Between S and S°

qrev = -w = RT lnVf

Vi

ΔS = qrev

T= R ln

Vf

Vi

ΔS = Sf – Si = R lnVf

Vi

= R lnPi

Pf

= -R lnPf

Pi

Sf = S°- R lnP

P°= S°- R ln

P1

= S° - R ln P

Let i be the standard state

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

SNH3= S°NH3

– Rln PNH3

SN2= S°N2

– Rln PN2

SH2 = S°H2

– Rln PH2

ΔSrxn = 2(S°NH3 – Rln PNH3

) – 2(S°N2 – Rln PN2

) –3(S°H2 – Rln PH2

)

ΔSrxn = 2 S°NH3 – 2S°N2

–3S°H2+ Rln

PN2PH2

PNH3

2

2

3

ΔSrxn = ΔS°rxn + RlnPN2

PH2

PNH3

2

2

3

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ΔG Under Non-standard Conditions

ΔG = ΔH° - TΔS ΔSrxn = ΔS°rxn + RlnPN2

PH2

PNH3

2

2

3

ΔG = ΔH° - TΔS°rxn – TR lnPN2

PH2

PNH3

2

2

3

ΔG = ΔG° + RT lnPN2

PH22

PNH3

2

3

ΔG = ΔG° + RT ln Q