Thermodynamics: Spontaneity, Entropy and Free Energy

ThermodynamicsThermodynamics::

Spontaneity, Spontaneity, Entropy and Free Entropy and Free

EnergyEnergy

SpontaneitySpontaneityA A spontaneousspontaneous process is one that occurs process is one that occurs without outside intervention. Examples without outside intervention. Examples include:include:- a ball rolling downhill- a ball rolling downhill- ice melting at temperatures above 0- ice melting at temperatures above 0ooCC- gases expanding to fill their container- gases expanding to fill their container- iron rusts in the presence of air and - iron rusts in the presence of air and waterwater- two gases mixing- two gases mixing

SpontaneitySpontaneitySpontaneous processes can Spontaneous processes can

release energy (a ball rolling release energy (a ball rolling downhill), require energy (ice downhill), require energy (ice melting at temperatures above melting at temperatures above 00ooC), or involve no energy change C), or involve no energy change at all (two gases mixing) . at all (two gases mixing) .

SpontaneitySpontaneityThere are three factors that There are three factors that

combine to predict spontaneity. combine to predict spontaneity. They are:They are:

1. Energy Change1. Energy Change2. Temperature2. Temperature3. Entropy Change3. Entropy Change

EntropyEntropy

A measure A measure of of

randomness randomness or disorderor disorder

EntropyEntropyEntropy, S, is a measure of Entropy, S, is a measure of

randomness or disorder. The randomness or disorder. The natural tendency of things is to tend natural tendency of things is to tend toward greater disorder. This is toward greater disorder. This is because there are many ways (or because there are many ways (or positions) that lead to disorder, but positions) that lead to disorder, but very few that lead to an ordered very few that lead to an ordered state.state.

EntropyEntropy

The driving force for a The driving force for a spontaneous process is an spontaneous process is an

increase in the entropy of increase in the entropy of the the universeuniverse..

EntropyEntropy

ΔΔSSo o and Phase Changesand Phase Changes

Gases have more entropy than liquids or solids.

ΔΔSSo o and Mixturesand Mixtures

Mixtures have more entropy than pure substances.

Entropy Values of Entropy Values of Common SubstancesCommon Substances

The 2The 2ndnd Law of Law of ThermodynamicsThermodynamics

In any spontaneous process there In any spontaneous process there is always is always

an increase in the entropy of the an increase in the entropy of the universe.universe.


Water spontaneously freezes at a Water spontaneously freezes at a temperature below 0temperature below 0ooC. Therefore, the C. Therefore, the process increases the entropy of process increases the entropy of the the universeuniverse. . The water molecules become much more The water molecules become much more ordered as they freeze, and experience a ordered as they freeze, and experience a decrease in entropy. The process also decrease in entropy. The process also releases heat, and this heat warms releases heat, and this heat warms gaseous molecules in air, and increases gaseous molecules in air, and increases the entropy of the surroundings.the entropy of the surroundings.


Since the process is Since the process is spontaneous below 0spontaneous below 0ooC, C, ΔΔSSsurrsurr, , which is positive, must be greater which is positive, must be greater in magnitude than in magnitude than ΔΔS of the water S of the water molecules.molecules.

ΔΔ S and Spontaneity S and Spontaneity

SpontaneitySpontaneityEntropy, temperature and heat Entropy, temperature and heat

flow all play a role in spontaneity. A flow all play a role in spontaneity. A thermodynamic quantity, the thermodynamic quantity, the Gibbs Gibbs Free EnergyFree Energy (G), combines these (G), combines these factors to predict the spontaneity of factors to predict the spontaneity of a process.a process.

ΔΔG = G = ΔΔH - TH - TΔΔS S

SpontaneitySpontaneityΔΔG = G = ΔΔH - TH - TΔΔS S

If a process releases heat (If a process releases heat (ΔΔH is H is negative) and has an increase in negative) and has an increase in entropy (entropy (ΔΔS is positive), it will S is positive), it will always be spontaneous. always be spontaneous.

The value of The value of ΔΔG for spontaneous G for spontaneous processes is negative.processes is negative.

SpontaneitySpontaneityΔΔG = G = ΔΔH - TH - TΔΔSS

Spontaneity and Spontaneity and ΔΔGGIf If ΔΔG is negative, the process is G is negative, the process is

spontaneous (and the reverse spontaneous (and the reverse process is non-spontaneous).process is non-spontaneous).

If If ΔΔG is positive, the process is G is positive, the process is non-spontaneous, and the reverse non-spontaneous, and the reverse process is spontaneous.process is spontaneous.

If If ΔΔG = 0, the system is at G = 0, the system is at equilibrium. equilibrium.

ΔΔGGAlthough Although ΔΔG can be used to predict in G can be used to predict in which direction a reaction will proceed, it which direction a reaction will proceed, it does not predict the rate of the reaction. does not predict the rate of the reaction. For example, the conversion of diamond For example, the conversion of diamond to graphite has a to graphite has a ΔΔGGoo = -3 kJ, so = -3 kJ, so diamonds should spontaneously change diamonds should spontaneously change to graphite at standard conditions. to graphite at standard conditions. However, kinetics shows that the reaction However, kinetics shows that the reaction is extremely slow.is extremely slow.

The Significance of The Significance of ΔΔGGΔΔG represents the driving force G represents the driving force

for the reaction to proceed to for the reaction to proceed to equilibrium.equilibrium.

The Significance of The Significance of ΔΔGGIf negative, the value of If negative, the value of ΔΔG in KJ is the G in KJ is the maximum possible useful work that can maximum possible useful work that can be obtained from a process or reaction be obtained from a process or reaction at constant temperature and pressure.at constant temperature and pressure.

If positive, the value of If positive, the value of ΔΔG in KJ is the G in KJ is the minimum work that must be done to minimum work that must be done to make the non-spontaneous process or make the non-spontaneous process or reaction proceed.reaction proceed.

Predicting the sign of Predicting the sign of ΔΔSSoo

For many chemical reactions or For many chemical reactions or physical changes, it is relatively easy physical changes, it is relatively easy to predict if the entropy of the to predict if the entropy of the system is increasing or decreasing.system is increasing or decreasing.

If a substance goes from a more If a substance goes from a more ordered phase (solid) to a less ordered phase (solid) to a less ordered phase (liquid or gas), its ordered phase (liquid or gas), its entropy increases.entropy increases.

Predicting the sign of Predicting the sign of ΔΔSSoo

For chemical reactions, it is For chemical reactions, it is sometimes possible to compare the sometimes possible to compare the randomness of products versus randomness of products versus reactants.reactants.

2 KClO2 KClO33((ss) ) 2 KCl(s) + 3 O 2 KCl(s) + 3 O22((gg))The production of a gaseous product The production of a gaseous product from a solid reactant will have a from a solid reactant will have a positive value of positive value of ΔΔSSoo..

Calculating Entropy Calculating Entropy ChangesChanges

Since entropy is a measure of Since entropy is a measure of randomness, it is possible to calculate randomness, it is possible to calculate absolute entropy values. This is in absolute entropy values. This is in contrast to enthalpy values, where we contrast to enthalpy values, where we can only calculate can only calculate changeschanges in enthalpy. in enthalpy. A perfect crystal at absolute zero has an A perfect crystal at absolute zero has an entropy value (S) =0. All other entropy value (S) =0. All other substances have positive values of substances have positive values of entropy due to some degree of disorder.entropy due to some degree of disorder.


Fortunately, the entropy values Fortunately, the entropy values of most common elements and of most common elements and compounds have been tabulated. compounds have been tabulated. Most thermodynamic tables, Most thermodynamic tables, including the appendix in the including the appendix in the textbook, include standard entropy textbook, include standard entropy values, Svalues, Soo..

Entropy Values of Entropy Values of Common SubstancesCommon Substances


For any chemical reaction, For any chemical reaction,

Δ Δ SSooreactionreaction= = ΣΣmolmolprod prod SSoo

productsproducts- - ΣΣmolmolreact react

SSooreactantsreactants

The units of entropy are joules/K-mol.The units of entropy are joules/K-mol.

Calculation of ∆GCalculation of ∆Goo

∆∆GGoo, the standard free energy , the standard free energy change, can be calculated in several change, can be calculated in several ways.ways.

∆∆GGo o = ∆H= ∆Hoo - T ∆S - T ∆Soo

It can be calculated directly, using the It can be calculated directly, using the standard enthalpy change and entropy standard enthalpy change and entropy change for the process.change for the process.



∆ ∆HHoo is usually calculated by using is usually calculated by using standard enthalpies of formation, standard enthalpies of formation, ∆H∆Hff

oo..

∆∆HHoorxnrxn = = ΣΣmolmolprod prod ∆H∆Hoo

productsproducts- - ΣΣmolmolreact react

∆H∆Hooreactantsreactants



Once ∆HOnce ∆Hoo and ∆S and ∆Soo have been have been calculated, the value of ∆Gcalculated, the value of ∆Goo can be can be calculated, using the temperature in calculated, using the temperature in Kelvins.Kelvins.


∆ ∆GGo o can also be calculated by can also be calculated by combining the free energy changes combining the free energy changes of related reactions. This is the of related reactions. This is the same method used in Hess’ Law to same method used in Hess’ Law to calculate enthalpy changes. If the calculate enthalpy changes. If the sum of the reactions gives the sum of the reactions gives the reaction of interest, the sum of the reaction of interest, the sum of the ∆G∆Go o values gives ∆Gvalues gives ∆Go o for the for the reaction.reaction.


Lastly, ∆GLastly, ∆Goo can be calculated can be calculated using standard free energies of using standard free energies of formation, ∆Gformation, ∆Gff

oo. Some tables of . Some tables of thermodynamic data, including the thermodynamic data, including the appendix of your textbook, include appendix of your textbook, include values of ∆Gvalues of ∆Gff

oo. .

∆∆GGrxnrxno o = = ΣΣmolmolprod prod ∆G∆Gff

o o prod prod - - ΣΣmolmolreact react ∆G∆Gff

o o reactreact


When calculating ∆GWhen calculating ∆Go o from from standard free energies of formation, standard free energies of formation, keep in mind that ∆Gkeep in mind that ∆Gff

o o for any for any element in its standard state is zero. element in its standard state is zero. As with enthalpies of formation, the As with enthalpies of formation, the formation reaction is the reaction of formation reaction is the reaction of elements in their standard states to elements in their standard states to make compounds (or allotropes). make compounds (or allotropes).



Note the values of zero for nitrogen, hydrogen and graphite.

Spontaneity ProblemSpontaneity Problem Consider the reaction: Consider the reaction:

CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + CO) + CO22((gg) at 25) at 25ooC. C. Calculate ∆GCalculate ∆Goo using the tables in the using the tables in the appendix of your textbook. Is the process appendix of your textbook. Is the process spontaneous at this temperature? Is it spontaneous at this temperature? Is it spontaneous at all temperatures? If not, spontaneous at all temperatures? If not, at what temperature does it become at what temperature does it become spontaneous? spontaneous?


CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + CO) + CO22((gg) at 25) at 25ooC. C. Calculate ∆GCalculate ∆Goo using the tables in the using the tables in the appendix of your textbook. Is the process appendix of your textbook. Is the process spontaneous at this temperature? spontaneous at this temperature? Calculation of ∆GCalculation of ∆Grxnrxn

oo will indicate will indicate spontaneity at 25spontaneity at 25ooC. It can be calculated C. It can be calculated using using ∆∆GGff

oo values or from ∆H values or from ∆Hffoo and ∆S and ∆Soo values. values.


CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + CO) + CO22((gg)) ∆ ∆GGrxnrxn

o o = = ΣΣnnprod prod ∆G∆Gffo o

prod prod - - ΣΣnnreact react ∆G∆Gffo o

reactreact


CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + CO) + CO22((gg)) ∆∆GGrxnrxn

o o =[(1 mol) (-604.0 kJ/mol) + (1 mol)(-=[(1 mol) (-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] 394.4 kJ/mol)] –[1 mol(-1128.8 kJ/mol)] –[1 mol(-1128.8 kJ/mol)]


CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + CO) + CO22((gg)) ∆∆GGrxnrxn

o o =[(1 mol) (-604.0 kJ/mol) + (1 mol)(-=[(1 mol) (-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)] 394.4 kJ/mol)] –[1 mol(-1128.8 kJ/mol)] = –[1 mol(-1128.8 kJ/mol)] = +130.4 kJ+130.4 kJ


CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + CO) + CO22((gg) at 25) at 25ooC. C. Calculate ∆GCalculate ∆Goo using the tables in the using the tables in the appendix of your textbook. Is the process appendix of your textbook. Is the process spontaneous at this temperature? spontaneous at this temperature? Since ∆GSince ∆Grxnrxn

o o =+130.4 kJ, the reaction is =+130.4 kJ, the reaction is not spontaneousnot spontaneous at 25 at 25ooC.C.

Spontaneity ProblemSpontaneity Problem Consider the reaction: Consider the reaction: CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + CO) + CO22((gg) at 25) at 25ooC. C.

Is it spontaneous at all Is it spontaneous at all temperatures? If not, at what temperatures? If not, at what temperature does it become temperature does it become spontaneous? spontaneous?


CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + CO) + CO22((gg) at 25) at 25ooC. C. Is it spontaneous at all temperatures? If Is it spontaneous at all temperatures? If not, at what temperature does it become not, at what temperature does it become spontaneous?spontaneous?

At 25At 25ooC, ∆GC, ∆Grxnrxnoo is positive, and the is positive, and the

reaction is not spontaneous in the reaction is not spontaneous in the forward direction. forward direction.


CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + CO) + CO22((gg) at 25) at 25ooC. C. Is it spontaneous at all temperatures? Is it spontaneous at all temperatures? If not, at what temperature does it If not, at what temperature does it become spontaneous?become spontaneous?Inspection of the reaction shows that it Inspection of the reaction shows that it involves an involves an increaseincrease in entropy due to in entropy due to production of a gas from a solid.production of a gas from a solid.


CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + CO) + CO22((gg) at 25) at 25ooC. C. Is it spontaneous at all temperatures? Is it spontaneous at all temperatures? If not, at what temperature does it If not, at what temperature does it become spontaneous?become spontaneous?We can calculate the entropy change We can calculate the entropy change and the enthalpy change, and then and the enthalpy change, and then determine the temperature at which determine the temperature at which spontaneity will occur.spontaneity will occur.

CaCOCaCO33((ss) ↔CaO() ↔CaO(ss) + ) + COCO22((gg))

Since ∆GSince ∆Go o = ∆H= ∆Hoo - T∆S - T∆Soo, and , and there is an increase in entropy, the there is an increase in entropy, the reaction will become spontaneous at reaction will become spontaneous at higher temperatures.higher temperatures.

To calculate ∆STo calculate ∆Soo, use the , use the thermodynamic tables in the thermodynamic tables in the appendix.appendix.


∆Srxno =[1mol(213.6J/K-mol)

+1mol(39.7J/K-mol)]-[1mol(92.9J/K-mol)] = 160.4

J/K


∆ ∆GGo o = ∆H= ∆Hoo - T∆S - T∆So o

Since we know the value of ∆GSince we know the value of ∆Go o

(+130.4 kJ) and ∆S(+130.4 kJ) and ∆So o ((160.4 J/K), we can calculate the value of ∆H∆Hoo at 25 at 25ooC.C.

130.4 kJ = ∆H130.4 kJ = ∆Hoo –(298K) ( –(298K) (160.4 J/K)∆∆HHo o = 130.4 kJ + (298K) (.= 130.4 kJ + (298K) (.1604 kJ/K)

∆∆HHo o = + 178.2 kJ= + 178.2 kJ



If we assume that the values of If we assume that the values of ∆H∆Hoo and ∆S and ∆So o don’t change much with don’t change much with temperature, we can estimate the temperature, we can estimate the temperature at which the reaction temperature at which the reaction will become spontaneous.will become spontaneous.



∆∆GGo o is positive at lower temperatures, is positive at lower temperatures, and will be negative at higher and will be negative at higher temperatures. Set ∆Gtemperatures. Set ∆Go o equal to zero, equal to zero, and solve for temperature.and solve for temperature.

0 = ∆H∆Hoo - T∆S - T∆So o

T = T = ∆H∆Hoo ∆ ∆SSoo



0 = ∆H∆Hoo - T∆S - T∆So o

T = T = ∆H∆Hoo ∆ ∆SSoo

T = (178.2 kJ)/(T = (178.2 kJ)/(160.4 J/K)(10-3kJ/J)=1111K or 838oC

The reaction will be spontaneous in the forward direction at temperatures above 838oC.

∆∆G for Non-Standard G for Non-Standard ConditionsConditions

The thermodynamic tables are The thermodynamic tables are for standard conditions. This for standard conditions. This includes having all reactants includes having all reactants and and productsproducts present initially at a present initially at a temperature of 25temperature of 25ooC. All gases are C. All gases are at a pressure of 1 atm, and all at a pressure of 1 atm, and all solutions are 1 M.solutions are 1 M.


For non-standard temperature, For non-standard temperature, concentrations or gas pressures: concentrations or gas pressures:

∆∆G = ∆GG = ∆Goo + RTlnQ + RTlnQ

Where R = 8.314 J/K-molWhere R = 8.314 J/K-molT is temperature in KelvinsT is temperature in KelvinsQ is the reaction quotientQ is the reaction quotient


For non-standard temperature, For non-standard temperature, concentrations or gas pressures: concentrations or gas pressures:

∆∆G = ∆GG = ∆Goo + RTlnQ + RTlnQ

For Q, gas pressures are in For Q, gas pressures are in atmospheres, and concentrations of atmospheres, and concentrations of

solutions are in molarity, M.solutions are in molarity, M.

∆∆GGo o and Equilibrium and EquilibriumA large negative value of ∆GA large negative value of ∆Goo

indicates that the forward reaction indicates that the forward reaction or process is spontaneous. That is, or process is spontaneous. That is, there is a large there is a large driving forcedriving force for the for the forward reaction. This also means forward reaction. This also means that the equilibrium constant for the that the equilibrium constant for the reaction will be large. reaction will be large.

∆∆GGo o and Equilibrium and EquilibriumA large positive value of ∆GA large positive value of ∆Goo indicates indicates that the reverse reaction or process that the reverse reaction or process is spontaneous. That is, there is a is spontaneous. That is, there is a large large driving forcedriving force for the reverse for the reverse reaction. This also means that the reaction. This also means that the equilibrium constant for the reaction equilibrium constant for the reaction will be small. will be small. When a reaction or process is at When a reaction or process is at equilibrium, ∆Gequilibrium, ∆Goo = zero. = zero.

∆∆GGo o and Equilibrium and Equilibrium

∆∆GGo o and Equilibrium and Equilibrium∆∆G = ∆GG = ∆Goo + RT lnQ + RT lnQ

At equilibrium, ∆G is equal to At equilibrium, ∆G is equal to zero, and zero, and

Q = K.Q = K.0 = ∆G0 = ∆Goo + RT lnK + RT lnK

∆∆GGoo = - RT lnK = - RT lnK

∆∆GGo o and Equilibrium and Equilibrium Calculate, ∆GCalculate, ∆Go o and K at 25and K at 25ooC for:C for:

C (s, C (s, diamonddiamond) ↔ C (s, ) ↔ C (s, graphitegraphite) )


C (s, C (s, diamonddiamond) ↔ C (s, ) ↔ C (s, graphitegraphite))∆∆GGo o = (1 mol) ∆G= (1 mol) ∆Goo

f (graphite)f (graphite) - (1 mol) ∆G - (1 mol) ∆Goof f

(diamond)(diamond)

= 0 -(1 mol)(2.900 kJ/mol) = 0 -(1 mol)(2.900 kJ/mol) = -2.900 kJ= -2.900 kJ

The reaction is spontaneous at 25The reaction is spontaneous at 25ooC.C.


C (s, C (s, diamonddiamond) ↔ C (s, ) ↔ C (s, graphitegraphite))∆∆GGo o = -2.900 kJ= -2.900 kJ

∆∆GGo o = -2.900 kJ = -RT ln K= -2.900 kJ = -RT ln K-2.900 kJ = -(8.314J/mol-K) (298.2K)ln -2.900 kJ = -(8.314J/mol-K) (298.2K)ln

KKln K = 1.170ln K = 1.170

K= eK= e1.1701.170 = 3.22 = 3.22

C (s, C (s, diamonddiamond) ↔ C (s, ) ↔ C (s, graphitegraphite))

The negative value of ∆GThe negative value of ∆Go o and the and the equilibrium constant >1 suggest that equilibrium constant >1 suggest that diamonds can spontaneously react to diamonds can spontaneously react to form graphite. form graphite. Although the reaction Although the reaction is is thermodynamically favoredthermodynamically favored, the rate , the rate constant is extremely small due to a constant is extremely small due to a huge activation energy. The disruption huge activation energy. The disruption of the bonding in the diamond to form of the bonding in the diamond to form planar spplanar sp2 2 hybridized carbon atoms is hybridized carbon atoms is kinetically unfavorable.kinetically unfavorable.

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Thermodynamics: Spontaneity, Entropy and Free Energy