Chapter 20Chapter 20Thermodynamics and Thermodynamics and

EquilibriumEquilibrium

OverviewOverview

First Law of ThermodynamicsFirst Law of Thermodynamics

Spontaneous Processes and Spontaneous Processes and EntropyEntropy Entropy and the Second Law of Entropy and the Second Law of

ThermodynamicsThermodynamics Standard Entropies and the Third Law Standard Entropies and the Third Law

of Thermodynamicsof Thermodynamics

Free Energy ConceptFree Energy Concept Free Energy and SpontaneityFree Energy and Spontaneity Interpretation of Free EnergyInterpretation of Free Energy

Free Energy and Equilibrium Free Energy and Equilibrium ConstantsConstants Relating Relating G° to the Equilibrium ConstantG° to the Equilibrium Constant Change of Free Energy with Change of Free Energy with

TemperatureTemperature

DefinitionsDefinitions

Spontaneous or Product-favored reactionSpontaneous or Product-favored reaction: : reaction in which most of the reactants reaction in which most of the reactants can eventually be converted to products, can eventually be converted to products, given sufficient timegiven sufficient time

Nonspontaneous or Reactant-favored Nonspontaneous or Reactant-favored reactionreaction: misleading - does not mean that : misleading - does not mean that it does not occur at all, rather, it means it does not occur at all, rather, it means that when equilibrium is achieved, not that when equilibrium is achieved, not many reactant molecules have been many reactant molecules have been converted into products.converted into products.

Definitions ContinuedDefinitions Continued

ThermodynamicsThermodynamics: the science of : the science of energy transfer, it helps us to predict energy transfer, it helps us to predict whether a reaction can occur given whether a reaction can occur given enough time. Thermodynamics tells enough time. Thermodynamics tells us nothing about the speed of the us nothing about the speed of the reactions.reactions.

Reaction ProbabilityReaction Probability

After an exothermic reaction, energy is After an exothermic reaction, energy is distributed more randomly - dispersed over distributed more randomly - dispersed over a much larger number of atoms and a much larger number of atoms and molecules - than it was before. Energy molecules - than it was before. Energy dispersal is favored because it is much more dispersal is favored because it is much more probable that energy will be dispersed than probable that energy will be dispersed than that it will be concentrated.that it will be concentrated.

Just as there is a tendency for highly Just as there is a tendency for highly concentrated energy to disperse, concentrated energy to disperse, highly concentrated matter also tends highly concentrated matter also tends to disperse.to disperse.

There are two ways that the final state of a There are two ways that the final state of a system can be more probable than the initial system can be more probable than the initial oneone

1. Having energy dispersed over a 1. Having energy dispersed over a greater number of atoms and greater number of atoms and molecules andmolecules and

2. Having the atoms and molecules 2. Having the atoms and molecules themselves more disorderedthemselves more disordered

Entropy: A Measure of Matter Entropy: A Measure of Matter Dispersal or DisorderDispersal or Disorder

The dispersal or disorder in sample of The dispersal or disorder in sample of matter can be measured with a matter can be measured with a calorimeter, the same instrument calorimeter, the same instrument needed to measure the enthalpy needed to measure the enthalpy change when a reaction occurs. change when a reaction occurs.

The result is a thermodynamic The result is a thermodynamic function called entropy and function called entropy and symbolized by S. symbolized by S.

Entropy and the Third LawEntropy and the Third Law

Measurement of entropy depends on Measurement of entropy depends on the assumption that in a perfect the assumption that in a perfect crystal at the absolute zero crystal at the absolute zero temperature all translational motion temperature all translational motion ceases and there is not any disorder.ceases and there is not any disorder.

Calculating Entropy ChangeCalculating Entropy Change

When energy is transferred to matter When energy is transferred to matter in very small increments, so that the in very small increments, so that the temperature change is very small, temperature change is very small, the entropy change can be the entropy change can be calculated as calculated as S = q/TS = q/T, the heat , the heat absorbed divided by the absolute absorbed divided by the absolute temperature at which the change temperature at which the change occurs.occurs.

Standard Molar EntropiesStandard Molar Entropies

Applies to one mole of a substance at Applies to one mole of a substance at standard pressure. Expressed in standard pressure. Expressed in units of joules per mole*kelvin.units of joules per mole*kelvin.

S = Sproducts – SreactantsS = Sproducts – Sreactants

Example 1Example 1

Calculate the standard molar entropy Calculate the standard molar entropy change for the formation of gaseous change for the formation of gaseous propane.propane.

SSoocc = 6 J/K = 6 J/K

SSooHH = 131 J/K = 131 J/K

SSooC3H8C3H8 = 207 J/K = 207 J/K

Generalizations About EntropyGeneralizations About Entropy

1. When comparing the same or very similar 1. When comparing the same or very similar substances, entropies of gases are much larger substances, entropies of gases are much larger than those of liquids, which are larger than for than those of liquids, which are larger than for solids.solids.

2. Entropies of more complex molecules are larger 2. Entropies of more complex molecules are larger than those of simpler molecules, especially in a than those of simpler molecules, especially in a series of closely related compounds.series of closely related compounds.

3. Entropies of ionic solids become larger as the 3. Entropies of ionic solids become larger as the attractions among the ions become weaker.attractions among the ions become weaker.

4. Entropy usually increases when a pure liquid or 4. Entropy usually increases when a pure liquid or solid dissolves in a solution.solid dissolves in a solution.

5. Entropy increases when a dissolved gas escapes 5. Entropy increases when a dissolved gas escapes from a solution.from a solution.

Example 2Example 2 Indicate whether the entropy increases or Indicate whether the entropy increases or

decreases in each of the following decreases in each of the following processes.processes.

1.1. Moisture condenses on the outside of cold Moisture condenses on the outside of cold glassglass

2.2. Gasoline vaporizes in the carburetor of an Gasoline vaporizes in the carburetor of an automobile engineautomobile engine

3.3. Sugar dissolves in coffeeSugar dissolves in coffee4.4. Iron rustsIron rusts

Example 3Example 3

Predict the sign of Predict the sign of S for the S for the following:following:

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

2.2. 2H2H22(g) + O(g) + O22(g) (g) 2H 2H22O(l)O(l)

3.3. HH22O(l) O(l) H H22O(g)O(g)

4.4. NN22(g) + 3H(g) + 3H22 (g) (g) 2NH 2NH33(g)(g)

Entropy and the Second Law of Entropy and the Second Law of ThermodynamicsThermodynamics

In a product-favored process there is a net In a product-favored process there is a net increase in the entropy of the system and increase in the entropy of the system and the surroundings. the surroundings.

In other words In other words when a product favored when a product favored reaction occurs the entropy (disorder) reaction occurs the entropy (disorder) of the universe increasesof the universe increases. This means . This means that even if the entropy of a particular that even if the entropy of a particular system decreases in a product favored system decreases in a product favored process, the total change in the entropy of process, the total change in the entropy of the universe (the system and all its the universe (the system and all its surroundings) must be positive. surroundings) must be positive.

Entropy of the SurroundingsEntropy of the Surroundings

The sign of The sign of Ssurr depends on Ssurr depends on the direction of heat flow (q).the direction of heat flow (q).

A positive A positive Ssurr occurs when Ssurr occurs when heat flows out of the system into heat flows out of the system into the surroundings and increases the surroundings and increases thermal motion. thermal motion. This should make This should make sense to you because heat is a form sense to you because heat is a form of energy and when the energy of energy and when the energy increases the molecular motion increases the molecular motion increases which causes more increases which causes more randomness. randomness.

A negative A negative Ssurr occurs when Ssurr occurs when heat flows into the system from heat flows into the system from the surroundings, decreasing the the surroundings, decreasing the thermal motion of the thermal motion of the surroundings and therefore surroundings and therefore decreasing the entropy.decreasing the entropy.

Ssurr and absolute temperatureSsurr and absolute temperature

If the surroundings are at a high If the surroundings are at a high temperature, the various types of temperature, the various types of molecular motion are already molecular motion are already sufficiently energetic. sufficiently energetic.

Therefore, the absorption of heat Therefore, the absorption of heat from an exothermic process in the from an exothermic process in the system will have relatively little system will have relatively little impact on molecular motions and the impact on molecular motions and the resulting increase in entropy will be resulting increase in entropy will be small. small.

If the temperature of the surroundings If the temperature of the surroundings is low, then the addition of the same is low, then the addition of the same amount of heat will cause a more amount of heat will cause a more drastic increase in molecular motions drastic increase in molecular motions and hence a larger increase in entropy. and hence a larger increase in entropy.

So, the entropy change produced when So, the entropy change produced when a given amount of heat is transferred is a given amount of heat is transferred is greater at low temperatures than at greater at low temperatures than at high temperatures. high temperatures.

The result is that The result is that Ssurr = qsurr/T ; Ssurr = qsurr/T ; where qsurr is the heat flow into the where qsurr is the heat flow into the surroundings at the absolute (Kelvin) surroundings at the absolute (Kelvin) temperature, T. temperature, T.

For a constant-pressure processes For a constant-pressure processes H H was defined as being equal to the was defined as being equal to the qsystem which would be equal to -qsystem which would be equal to -qsurr. qsurr.

So this means that So this means that Ssurr = -Ssurr = -H/TH/T

S and S and HH

For a product-favored process:For a product-favored process:

Ssystem - Ssystem - H/T > 0 orH/T > 0 or

TTS - S - H > 0H > 0

Second Law of ThermodynamicsSecond Law of Thermodynamics

To be product favored a reaction must lead to To be product favored a reaction must lead to an increase in the entropy of the universe. an increase in the entropy of the universe.

For a product favored process TFor a product favored process TS - S - H > 0 or H > 0 or if we multiply the equation throughout by -1: if we multiply the equation throughout by -1: H - TH - TS < 0. S < 0.

Now we have a criterion for a product-favored Now we have a criterion for a product-favored reaction that is expressed only in terms of the reaction that is expressed only in terms of the properties of the system and we no longer properties of the system and we no longer need be concerned with the surroundings. need be concerned with the surroundings.

Gibbs Free Energy Gibbs Free Energy

A new thermodynamic function can now A new thermodynamic function can now be introduced, its called the Gibbs Free be introduced, its called the Gibbs Free Energy, or simply Free Energy, as follows: Energy, or simply Free Energy, as follows:

G = H - TS G = H - TS G has the units of energy and, like H and G has the units of energy and, like H and

S, it is a state function. (State Function = S, it is a state function. (State Function = A quantity whose value is determined A quantity whose value is determined only by the state of the system)only by the state of the system)

GG

The change in free energy (The change in free energy (G) of a G) of a system (which if what we're system (which if what we're interested in) for a process at interested in) for a process at constant temperature is given by constant temperature is given by G G = = H - TH - TSS

Example 4Example 4

Given the values of Given the values of H and H and S, which S, which of the following changes will be of the following changes will be spontaneous at constant T and P?spontaneous at constant T and P?

1.1. H= +25KJ; H= +25KJ; S = +5.0 J/K; T = 300 KS = +5.0 J/K; T = 300 K

2.2. H = +25 KJ; H = +25 KJ; S = +100J/K; T = 300 KS = +100J/K; T = 300 K

3.3. H = -10 KJ; H = -10 KJ; S = +5.0 J/K; T = 298 KS = +5.0 J/K; T = 298 K

4.4. H = -10 KJ; H = -10 KJ; S = -40.0 J/K; T = 200 KS = -40.0 J/K; T = 200 K

Standard Free Energies of Standard Free Energies of FormationFormation

The standard free energy of The standard free energy of formation, Gformation, Gff, of a substance is , of a substance is defined similarly to the standard defined similarly to the standard enthalpy of formation. That is, enthalpy of formation. That is, GGff is is the free-energy change that occurs the free-energy change that occurs when 1 mol of a substance is formed when 1 mol of a substance is formed from its elements in their most stable from its elements in their most stable states at 1 atm and at a specified states at 1 atm and at a specified temperature (usually 25 C)temperature (usually 25 C)

By tabulating By tabulating GGff for substances we can for substances we can easily calculate easily calculate G for any reaction involving G for any reaction involving those substances using the following formula:those substances using the following formula:

G = G = Gf (products) - Gf (products) - Gf (reactants)Gf (reactants)

NOTE: If you are confused about the two NOTE: If you are confused about the two different ways to calculate the different ways to calculate the G - this last G - this last one can only be used when the temperature one can only be used when the temperature is that of the tabulated values - usually 25 C.is that of the tabulated values - usually 25 C.

Example 5Example 5

For the following reaction at 298 K, For the following reaction at 298 K, determine the value of the Gibbs Free determine the value of the Gibbs Free Energy Energy

4PH4PH33(g) + 8O(g) + 8O22(g) (g) P P44OO1010(s) + 6H(s) + 6H22O(l)O(l)

GGf f PHPH3 3 = 13 KJ/mol= 13 KJ/mol

GGf f OO2 2 = 0 KJ/mol= 0 KJ/mol

GGf f PP44OO10 10 = -2698 KJ/mol= -2698 KJ/mol

GGf f HH22O = -237 KJ/molO = -237 KJ/mol

Product-Favored or Reactant-Product-Favored or Reactant-Favored?Favored?

Reactions at constant temperature and Reactions at constant temperature and pressure go in such a direction as to pressure go in such a direction as to decrease the free energy of the system.decrease the free energy of the system.

G < 0 G < 0 Product-Favored ReactionProduct-Favored Reaction

G > 0G > 0 Reactant-Favored. Reactant-Favored.

G = 0G = 0 The system is at The system is at equilibrium. equilibrium. There is no net There is no net change. change.

Free Energy and TemperatureFree Energy and Temperature

The value of G and consequently the The value of G and consequently the directionality of the reaction change directionality of the reaction change with temperature.with temperature.

Free-Energy and TemperatureFree-Energy and Temperature

HH SS Low TemperatureLow Temperature High High TemperatureTemperature

++ ++ reactant-favoredreactant-favored product-favoredproduct-favored

++ -- reactant-favoredreactant-favored reactant-favoredreactant-favored

-- ++ product-favoredproduct-favored product-product-favoredfavored

-- -- product-favoredproduct-favored reactant-reactant-favoredfavored

G – Temperature DependenceG – Temperature Dependence

SpontaneitySpontaneity: Reaction becomes : Reaction becomes spontaneous when spontaneous when G goes from + G goes from + to to . We use . We use G = 0 to tell us when G = 0 to tell us when reaction just becomes spontaneous reaction just becomes spontaneous or 0 = or 0 = H H T TS or T = S or T = H/H/S. S.

Example 6Example 6

Determine the temperature at which Determine the temperature at which the synthesis of HI(g) becomes the synthesis of HI(g) becomes spontaneous. spontaneous.

HHoo = +52.96 kJ and = +52.96 kJ and SSoo = +166.4 = +166.4 J/molJ/mol

HH22(g) + I(g) + I22(g) (g) 2HI(g) 2HI(g)

EQUILIBRIUM CONSTANTS EQUILIBRIUM CONSTANTS AND AND GG

Equilibrium constant for a reaction Equilibrium constant for a reaction aA + bB +... aA + bB +... mM + nN + mM + nN + ... is defined as ... is defined as

Tells how far to right reaction proceeds. Tells how far to right reaction proceeds. Large value Large value mostly products. mostly products. Small value Small value mostly reactants. mostly reactants.

At equilibrium this equation must always be At equilibrium this equation must always be obeyed no matter what relative amount of obeyed no matter what relative amount of reactant and was started with.reactant and was started with.

...][B][A

...][N][M =K

ba

nm

Thermodynamics and the Thermodynamics and the Equilibrium Constant Equilibrium Constant

G = G = GGoo + RT lnQ + RT lnQ

At equilibrium Q = K and At equilibrium Q = K and G = 0G = 0

GGoo = -RT lnK = -RT lnK where R is 8.31 J/mol where R is 8.31 J/mol KK

Example 7 Example 7

Chloroform, formerly used as an Chloroform, formerly used as an anesthetic, and now believed to be a anesthetic, and now believed to be a carcinogen has a heat of vaporization carcinogen has a heat of vaporization of 31.4 KJ/mol. The entropy of of 31.4 KJ/mol. The entropy of vaporization is 94.2 J/mol*K. At what vaporization is 94.2 J/mol*K. At what temperature would you expect temperature would you expect chloroform to boil (i.e. at what chloroform to boil (i.e. at what temperature will the liquid and temperature will the liquid and vapor be in equilibrium)?vapor be in equilibrium)?

Thermodynamics Thermodynamics

First Law: The total energy of the First Law: The total energy of the universe is constantuniverse is constant

Second Law: The total entropy of the Second Law: The total entropy of the universe is always increasinguniverse is always increasing

Third Law: The entropy of a pure, Third Law: The entropy of a pure, perfectly formed crystalline perfectly formed crystalline substance at absolute zero is zerosubstance at absolute zero is zero

Neither of the first two laws of Neither of the first two laws of thermodynamics has ever been thermodynamics has ever been or can be proven. However, or can be proven. However, there has never been a single, there has never been a single, concrete example showing concrete example showing otherwise.otherwise.