Spontaneity Spontaneity Entropy Entropy

and and Free EnergyFree Energy

WHAT DRIVES A REACTION TO BE WHAT DRIVES A REACTION TO BE SPONTANEOUS?SPONTANEOUS?

ENTHALPY (ENTHALPY (H)H)

heat contentheat content (exothermic reactions are generally (exothermic reactions are generally

favored)favored)

ENTROPY (S) – disorder of a system

(more disorder is favored) Nature tends toward chaos! Think about your room at the end of the week!

Spontaneous reactions are those that occur without outside intervention. They may occur fast OR slow (that is kinetics).

Some reactions are very fast (like combustion of hydrogen) other reactions are very slow (like graphite turning to diamond)

ENTROPYENTROPYThe Second LawThe Second Law of Thermodynamics of Thermodynamics

The universe is constantly The universe is constantly increasing disorder. increasing disorder.

Rudolph Clausius ‘discovered’ it Rudolph Clausius ‘discovered’ it and gave it its symbol.)and gave it its symbol.)

ENTROPYENTROPYThe Third Law of The Third Law of ThermodynamicsThermodynamics

The entropy of a perfect crystal at 0 The entropy of a perfect crystal at 0 K K

is zero.is zero.

not a lot of perfect crystals out there not a lot of perfect crystals out there so, entropy values are RARELY ever so, entropy values are RARELY ever zero—even elementszero—even elements

So what?So what?

This means the absolute entropy of This means the absolute entropy of a substance can then be a substance can then be determined at any temperature determined at any temperature higher than 0 K. higher than 0 K.

(Handy to know if you ever need to (Handy to know if you ever need to defend defend whywhy G & H for elements = G & H for elements = 0. . . BUT S does not!)0. . . BUT S does not!)

Predicting the entropy of a Predicting the entropy of a system based on physical system based on physical evidence:evidence:

The greater the disorder or randomness in a The greater the disorder or randomness in a system, the larger the entropy.system, the larger the entropy.

The entropy of a substance always increases The entropy of a substance always increases as it changes from solid to liquid to gas.as it changes from solid to liquid to gas.

When a pure solid or liquid dissolves in a When a pure solid or liquid dissolves in a solvent, the entropy of the substance solvent, the entropy of the substance increases (carbonates are an exception! --increases (carbonates are an exception! --they interact with water and actually bring they interact with water and actually bring MORE order to the system)MORE order to the system)

Predicting the entropy of a Predicting the entropy of a system based on physical system based on physical evidence:evidence:

When a gas molecule escapes from a When a gas molecule escapes from a solvent, the entropy increases.solvent, the entropy increases.

Entropy generally increases with Entropy generally increases with increasing molecular complexity increasing molecular complexity (crystal structure: KCl vs CaCl(crystal structure: KCl vs CaCl22) since ) since there are more there are more MOVING MOVING electrons!electrons!

Reactions increasing the number of Reactions increasing the number of moles of particles often increase moles of particles often increase entropy.entropy.

In general, the greater the number of arrangements, the higher the entropy of the system!

Exercise 2Exercise 2 Predicting Entropy Predicting Entropy Changes Changes Predict the sign of the entropy change Predict the sign of the entropy change

for each of the following processes.for each of the following processes.

A: Solid sugar is added to water to A: Solid sugar is added to water to form a solution.form a solution.

B: Iodine vapor condenses on a cold B: Iodine vapor condenses on a cold surface to form crystals.surface to form crystals.

Solution:Solution:

A: +∆A: +∆SSB: -∆B: -∆SS

Sample Problem ASample Problem A

Which of the following has the Which of the following has the largestlargest increase in entropy?increase in entropy?

a) COa) CO22(s) (s) CO CO22(g)(g) b) Hb) H22(g) + Cl(g) + Cl22(g) (g) 2 HCl(g) 2 HCl(g) c) KNOc) KNO33(s) (s) KNO KNO33(l)(l) d) Cd) C(diamond)(diamond) C C(graphite)(graphite)

Answer: Answer:

a)a) the substance changes from a the substance changes from a highly organized state to a more highly organized state to a more disorganized state.disorganized state.

Calculating Entropy Calculating Entropy from tables of from tables of standard values:standard values: Just the same as calculating the Just the same as calculating the

enthalpy earlier.enthalpy earlier.

SS rxnrxn = = SS (products)(products) - - SS (reactants)(reactants)

S is + when disorder increases (favored)S is + when disorder increases (favored)S is – when disorder decreasesS is – when disorder decreases

Units are usually J/K mol (not kJ ---tricky!)Units are usually J/K mol (not kJ ---tricky!)

BIG MAMMA, BIG MAMMA, verse 2 verse 2

Sample Problem B Sample Problem B

Calculate the entropy change at 25Calculate the entropy change at 25C, in C, in J/K for:J/K for:

2 SO2 SO22(g) + O(g) + O22(g) (g) 2 SO 2 SO33(g)(g)

Given the following data:Given the following data: SOSO22(g) (g) 248.1 J/K mol248.1 J/K mol OO22(g) (g) 205.3 J/K mol 205.3 J/K mol SOSO33(g)(g) 256.6 J/K mol 256.6 J/K mol

Solution:Solution:

Entropy change = -188.3 J/K Entropy change = -188.3 J/K

ENTROPY CHANGES FOR ENTROPY CHANGES FOR REVERSIBLE PHASE CHANGESREVERSIBLE PHASE CHANGES

(that’s a phase change at constant (that’s a phase change at constant temperature)temperature)

ΔS = ΔS = heat transferred heat transferred = = q q temperature at which change occurs Ttemperature at which change occurs T

**where the heat supplied (endothermic) (q **where the heat supplied (endothermic) (q > 0) or evolved (exothermic) (q < 0) is > 0) or evolved (exothermic) (q < 0) is divided by the temperature in Kelvindivided by the temperature in Kelvin

**It is important here to note if the **It is important here to note if the reaction is endothermic or exothermic. reaction is endothermic or exothermic. The actual significance of this is really The actual significance of this is really dependent on the temperature at which dependent on the temperature at which the process occurs. the process occurs.

(i.e., If you gave a millionaire $100 it (i.e., If you gave a millionaire $100 it would not make much difference in his would not make much difference in his happiness; if you gave a poor college happiness; if you gave a poor college student $100 it would create a totally student $100 it would create a totally different expression of happiness!) different expression of happiness!)

EX: water (l @ 100 EX: water (l @ 100 → → water (g @ 100)water (g @ 100) the entropy will increase.the entropy will increase.

Taking favored conditions into Taking favored conditions into consideration, consideration,

the equation above rearranges into: the equation above rearranges into: S = - S = - H H T T

Give signs to ΔH following exo/endo Give signs to ΔH following exo/endo guidelines! (If reaction is exo.; entropy of guidelines! (If reaction is exo.; entropy of system increases—makes sense!) system increases—makes sense!)

Exercise 4Exercise 4 Determining ∆S Determining ∆Ssurrsurr

In the metallurgy of In the metallurgy of antimony, the pure metal is antimony, the pure metal is recovered via different recovered via different reactions, depending on the reactions, depending on the composition of the ore. composition of the ore.

For example, iron is used to For example, iron is used to reduce antimony in sulfide reduce antimony in sulfide ores:ores:

SbSb22SS33(s)+3 Fe(s) (s)+3 Fe(s) 2 Sb(s)+3 2 Sb(s)+3 FeS(s)FeS(s)

∆ ∆H H = -125kJ= -125kJ

Carbon is used as the reducing Carbon is used as the reducing agent for oxide ores:agent for oxide ores:

SbSb44OO66(s)+6 C(s) (s)+6 C(s) 4 Sb(s)+6 4 Sb(s)+6 CO(g)CO(g)

∆ ∆HH = 778kJ = 778kJ

Calculate ∆Calculate ∆SSsurrsurr for each of for each of these reactions at 25these reactions at 25C C and 1 atm.and 1 atm.

SbSb22SS33(s)+3 Fe(s) (s)+3 Fe(s) 2 Sb(s)+3 2 Sb(s)+3

FeS(s)FeS(s)

SbSb44OO66(s)+6 C(s)(s)+6 C(s)4 Sb(s)+6 CO(g)4 Sb(s)+6 CO(g)

Solution:Solution:

∆∆SSsurrsurr = -2.61 × 103 J/K = -2.61 × 103 J/K

SUMMARYSUMMARY

ENTROPYENTROPY::

S = + MORE DISORDER S = + MORE DISORDER

(FAVORED CONDITION) (FAVORED CONDITION)

S = - MORE ORDER S = - MORE ORDER

Whether a reaction will occur Whether a reaction will occur spontaneously spontaneously may bemay be determined by looking at the determined by looking at the S of the universe. S of the universe. ΔS ΔS systemsystem + ΔS + ΔS surroundingssurroundings = ΔS = ΔS universeuniverse

IF ΔS IF ΔS universeuniverse is +, then reaction is is +, then reaction is spontaneousspontaneous

IF ΔS IF ΔS universeuniverse is -, then reaction is is -, then reaction is NONspontaneousNONspontaneous

Consider Consider

2 H2 H22 (g) + O (g) + O22 (g) (g) →→ H H22O (g) O (g) ignite & rxn is fast!ignite & rxn is fast!

ΔSΔSsystemsystem = -88.9J/K = -88.9J/K

Entropy declines Entropy declines (due mainly to 3 (due mainly to 3 → → 2 moles of gas!)2 moles of gas!)

. . . to confirm we need to . . . to confirm we need to know entropy of know entropy of surroundingssurroundingsΔSΔSsurroundingssurroundings = = q q surroundingssurroundings T T (this comes from ΔH (this comes from ΔH

calc.)calc.)

HHsystemsystem = - 483.6 kJ = - 483.6 kJ

First law of thermodynamics demands that this energy is transferred from the system to the surroundings so...

-ΔHsystem = ΔHsurroundings

OR

- (- 483.6 kJ) = + 483.6 kJ

ΔSΔS°°surroundingssurroundings = =

ΔHΔH°°surroundingssurroundings = = + 483.6 kJ+ 483.6 kJ TT 298 K 298 K = 1620 J/K= 1620 J/K

Now we can find ΔSNow we can find ΔS°°universeuniverse

ΔS ΔS systemsystem + ΔS + ΔS surroundingssurroundings = ΔS = ΔS universeuniverse

(-88.9 J/K) + (1620 J/K) = 1530 J/K(-88.9 J/K) + (1620 J/K) = 1530 J/K

Even though the entropy of the system Even though the entropy of the system declines, the entropy change for the declines, the entropy change for the surroundings is SOOO large that the surroundings is SOOO large that the overall change for the universe is overall change for the universe is positive.positive.

Bottom line: Bottom line:

A process is spontaneous in spite A process is spontaneous in spite of a negative entropy change as of a negative entropy change as long as it is extremely long as it is extremely exothermic. exothermic.

Sufficient exothermicity Sufficient exothermicity offsets system ordering.offsets system ordering.

FREE ENERGY FREE ENERGY Calculation of Gibb’s free energy Calculation of Gibb’s free energy

is what is what ultimatelyultimately decides decides whether a reaction is whether a reaction is spontaneous or not. spontaneous or not.

NEGATIVE NEGATIVE G’s are G’s are spontaneousspontaneous. .

G can be calculated one of G can be calculated one of several ways: several ways:

Big Mamma, verse 3:Big Mamma, verse 3: GG rxnrxn = = GG (products)(products) - - GG (reactants)(reactants)

This works the same way as enthalpy This works the same way as enthalpy and and

entropy from tables of standard entropy from tables of standard values! values!

Standard molar free energy Standard molar free energy of formationof formation

Same song, 3rd verse. Same song, 3rd verse.

ΔGΔG°°ff = 0 = 0

for elements in standard statefor elements in standard state

GRAND Daddy:GRAND Daddy:

G = G = H - TH - TSS

This puts together all information This puts together all information thus far! thus far!

By far, one of the most beneficial By far, one of the most beneficial equations to learn for AP exam! equations to learn for AP exam!

Hess’s law summation Hess’s law summation

Works same as Hess’s in the Works same as Hess’s in the enthalpy section—sum up enthalpy section—sum up equations using the guidelines as equations using the guidelines as before. before.

G = G = GG°° + RT ln (Q) + RT ln (Q) Define terms: Define terms: G = free energy not at standard G = free energy not at standard

conditions conditions GG = free energy at standard conditions = free energy at standard conditionsR = universal gas constant 8.3145 J/molR = universal gas constant 8.3145 J/molK K T = temp. in KelvinT = temp. in Kelvinln = natural log ln = natural log Q = reaction quotient: (for gases this is the partial Q = reaction quotient: (for gases this is the partial pressures of the products divided by the partial pressures of the products divided by the partial pressures of the reactants—all raised to the power pressures of the reactants—all raised to the power of their coefficients) of their coefficients)

Q = Q = [products][products] [reactants][reactants]

““RatLink”: RatLink”:

GG = -RTlnK = -RTlnK Terms: Terms: Basically the same as above --- however, Basically the same as above --- however,

here the system is athere the system is at equilibrium equilibrium, so , so G G = 0 and K represents the equilibrium = 0 and K represents the equilibrium constant under standard conditions.constant under standard conditions.

K = K = [products] [products] still raised to power of still raised to power of

coefficientscoefficients [reactants] [reactants]

““nFe”: nFe”: GG = - nFE = - nFE Remember this!! Remember this!! Terms: Terms: GG = just like above—standard free = just like above—standard free energy energy n = number of moles of electrons n = number of moles of electrons transferred (look at ½ transferred (look at ½

reactions)reactions) F = Faraday’s constant 96,485 F = Faraday’s constant 96,485 Coulombs/mole electrons Coulombs/mole electrons EE = standard voltage = standard voltage ** one volt = joule/coulomb**** one volt = joule/coulomb**

BIG MAMMA, verse 3: BIG MAMMA, verse 3:

GG rxnrxn = = GG (products)(products) - - GG (reactants)(reactants)

Sample Problem C: Sample Problem C:

Find the free energy of formation for the Find the free energy of formation for the oxidation of water to produce hydrogen oxidation of water to produce hydrogen peroxide.peroxide.

2 H2 H22O(l) + OO(l) + O22(g) (g) 2 H 2 H22OO22(l)(l)

Given the following information:Given the following information: GG°°ff

HH22O(l) O(l) -56.7 kcal/mol -56.7 kcal/mol OO22(g)(g) 0 kcal/mol 0 kcal/mol HH22OO22(l)(l) -27.2 kcal/mol -27.2 kcal/mol

Solution:Solution:

Free energy of formation = 59.0 Free energy of formation = 59.0 kcal/mol kcal/mol

Exercise 9 (GRAND Daddy: Exercise 9 (GRAND Daddy: G = G = H - H - TTS) S) Calculating ∆H Calculating ∆H, ∆S, ∆S, and ∆G, and ∆G

Consider the reaction:Consider the reaction: 2 SO2 SO22(g)+O(g)+O22(g)(g)2 SO2 SO33(g)(g)

carried out at 25carried out at 25C and 1 atm. C and 1 atm.

Calculate ∆HCalculate ∆H, ∆S, ∆S, and ∆G, and ∆G using the following data:using the following data:

Solution:Solution:

∆∆HH = -198 kJ = -198 kJ∆∆SS = -187 J/K = -187 J/K∆∆GG = -142 kJ = -142 kJ

Exercise 10 (Hess’s law of Exercise 10 (Hess’s law of summation)summation) Calculating ∆G Calculating ∆G

Using the following data (at 25Using the following data (at 25C):C):

CCdiamonddiamond(s)+O(s)+O22(g)(g)COCO22(g) ∆G(g) ∆G= -397 kJ = -397 kJ (16.5)(16.5)CCgraphitegraphite(s)+O(s)+O22(g)(g) CO CO22(g) ∆G(g) ∆G= -394 kJ = -394 kJ (16.6)(16.6)Calculate ∆GCalculate ∆G for the reaction: for the reaction:

CCdiamonddiamond(s)(s)CCgraphitegraphite(s)(s)

Solution:Solution:

∆∆GG = -3 kJ = -3 kJ

Exercise 13 (Exercise 13 (G = G = GG + RT + RT ln(Q)) ln(Q)) Calculating ∆G Calculating ∆G One method for synthesizing methanol One method for synthesizing methanol

(CH(CH33OH) involves reacting carbon OH) involves reacting carbon monoxide and hydrogen gases:monoxide and hydrogen gases:

CO(g)+2 HCO(g)+2 H22(g)(g)CHCH33OH(l)OH(l)

Calculate ∆G at 25Calculate ∆G at 25C for this reaction C for this reaction where carbon monoxide gas at 5.0 atm where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are and hydrogen gas at 3.0 atm are converted to liquid methanol.converted to liquid methanol.

Solution:Solution:

∆∆G at 25G at 25C = -38 kJ/mol rxn C = -38 kJ/mol rxn

Exercise 15 (“RatLink”: Exercise 15 (“RatLink”: GG = - = -RTlnK )RTlnK ) Free Energy and Equilibrium II Free Energy and Equilibrium II

The overall reaction for the The overall reaction for the corrosion (rusting) of iron by corrosion (rusting) of iron by oxygen isoxygen is

4 Fe(s) + 3 O4 Fe(s) + 3 O22(g) (g) ↔↔ 2 Fe 2 Fe22OO33(s) (s)

Using the following data, calculate the Using the following data, calculate the equilibrium constant for this reaction at equilibrium constant for this reaction at 2525C.C.

Solution:Solution:

K = 10K = 10261261

Gibb’s equation can also be used to Gibb’s equation can also be used to calculate the phase change calculate the phase change temperature of a substance. During temperature of a substance. During the phase change, there is an the phase change, there is an equilibrium between phases so the equilibrium between phases so the value of value of GG is zero. is zero.

Really just like what we did earlier in Really just like what we did earlier in this unit with enthalpy and entropy! this unit with enthalpy and entropy!

Sample Problem:Sample Problem:

Find the thermodynamic boiling Find the thermodynamic boiling point ofpoint of

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

Given the following information:Given the following information: Hvap = +44 kJHvap = +44 kJ

Svap = 118.8 J/KSvap = 118.8 J/K

Solution:Solution:

Thermodynamic boiling point = Thermodynamic boiling point = 370K 370K

SUMMARY OF FREE ENERGY:SUMMARY OF FREE ENERGY:

G = + NOT SPONTANEOUS G = + NOT SPONTANEOUS G = - SPONTANEOUS G = - SPONTANEOUS

Conditions of Conditions of G:G:

HH SS Result Result negneg pospos spontaneous at all tempsspontaneous at all tempspospos pospos spontaneous at high spontaneous at high temps temps

negneg negneg spontaneous at low tempsspontaneous at low tempspospos negneg not spontaneous, evernot spontaneous, ever

Relationship to K and E: Relationship to K and E:

G KG K E E 0 at equilibrium; K = 10 at equilibrium; K = 1 0 0 negativenegative >1, products favored >1, products favored

positive positive positivepositive <1, reactants favored <1, reactants favored

negative negative

Consider the reaction: Consider the reaction: A (g) A (g) B (g) B (g)

(a) The change in free energy to reach equilibrium, (a) The change in free energy to reach equilibrium, beginning with 1.0 mol of A(g) at PA = 2.0 atmbeginning with 1.0 mol of A(g) at PA = 2.0 atm

(b) The change in free energy to reach equilibrium, (b) The change in free energy to reach equilibrium, beginning with 1.0 mol of B(g) at PB = 2.0 atm.beginning with 1.0 mol of B(g) at PB = 2.0 atm.

(c) The free energy profile for A(g) (c) The free energy profile for A(g) B(g) in a system B(g) in a system containing 1.0 mol (A plus B) at PTOTAL = 2.0 atm. containing 1.0 mol (A plus B) at PTOTAL = 2.0 atm.

Each point on the curve corresponds Each point on the curve corresponds to the total free energy of the system to the total free energy of the system for a given combination of A and B.for a given combination of A and B.

The value of free energy is not only The value of free energy is not only useful for defining spontaneity, it is useful for defining spontaneity, it is also very useful in understanding the also very useful in understanding the maximum amount of work produced maximum amount of work produced or required by a system at constant or required by a system at constant temperature and pressure:temperature and pressure:

G = wG = wmaxmax

When the reaction is spontaneous, When the reaction is spontaneous, G is the G is the energy available to do work such as moving a energy available to do work such as moving a piston or flowing of electrons. When piston or flowing of electrons. When G is G is positive, and thus non-spontaneous, it represents positive, and thus non-spontaneous, it represents the amount of work needed to make the process the amount of work needed to make the process occur.occur.