Spontaneity, Entropy

and Free Energy

SPONTANEOUS PROCESSES AND ENTROPY

First Law“Energy can neither be created nor destroyed“.

The energy of the universe is constant. Enthalpy

Heat content – exothermic is favored. Spontaneous Processes

Processes that occur without outside intervention

Spontaneous processes may be fast or slowMany forms of combustion are fast.Conversion of diamond to graphite is slow.

ENTROPY (S) A measure of the randomness or

disorder. The driving force for a spontaneous

process is an increase in the entropy of the universe.

Entropy is a thermodynamic function describing the number of arrangements that are available to a system.

Nature proceeds toward the states that have the highest probabilities of existing.

SECOND LAW OF THERMODYNAMICS

"In any spontaneous process there is always an increase in the entropy of the universe"

"The entropy of the universe is increasing"

For a given change to be spontaneous, Suniverse must be positive

Suniv = Ssys + Ssurr

ENTROPYThe Third Law of Thermodynamics: the entropy of a perfect crystal at 0K

is zero. NOTE: not a lot of perfect crystals out there so, entropy values are RARELY ever zero—even elements. This means the absolute entropy of a substance can then be determined at any temperature higher than 0 K.

POSITIONAL ENTROPY The probability of occurrence of a

particular state depends on the number of ways (microstates) in which that arrangement can be achieved

Ssolid < Sliquid << Sgas

POSITIONAL ENTROPY

The fewer molecules there are, the fewer microstates there can be. Increasing particles can increase Entropy.

PREDICTING ENTROPY1. The greater the disorder or randomness

in a system, the larger the entropy.2. The entropy of a substance always

increases as it changes from solid to liquid to gas.

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

PREDICTING ENTROPY4. When a gas molecule escapes from

a solvent, the entropy increases.5. Entropy generally increases with

increasing molecular complexity (crystal structure: KCl vs CaCl2) since there are more MOVING electrons.

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

CALCULATING ENTROPY CHANGE IN A REACTION

Entropy is an extensive property (a function of the number of moles)

Generally, the more complex the molecule, the higher the standard entropy value

0 0 oreaction products reactantsp rS n S n S

IN GENERAL

The greater the number of arrangements, the higher the entropy of the system!

ΔS°rxn = ΣΔS°(products) - ΣΔS°(reactants)

S is + when disorder increases (favored) S is – when disorder decreases Units are usually J/K∙ mol (not kJ – pay

attention!)

PRACTICE ONEWhich of the following has the largest increase in entropy?a) CO2(s) → CO2(g)

b) H2(g) + Cl2(g) → 2 HCl(g)

c) KNO3(s) → KNO3(l)

d) C(diamond) → C(graphite)

PRACTICE TWO

Predict the sign of the entropy change for each of the following processes.A: Solid sugar is added to water to form a solution.

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

PRACTICE THREECalculate the entropy change at 25°C, in J/K for:2 SO2(g) + O2(g) → 2 SO3(g)

 Given the following data:SO2(g) 248.1 J/K ∙ molO2(g) 205.3 J/K ∙ molSO3(g) 256.6 J/K ∙ mol

REVIEW The driving force of a spontaneous

process is an increase in the entropy of the universe.

Entropy can be viewed as a measure of molecular randomness or disorder. It is also a thermodynamic function that describes the numbers of arrangements (positions and/or energy levels) that are available to a system existing at a given state.

Nature spontaneously proceeds towards the states that has the highest probability of existing.

REVIEW

Dissolving = > disorderS → L → G = > disorderEvaporation = >disorderBigger molecules = >

disorderReaction with more moles =

> disorder

POSITIONAL PROBABILITYPossible

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

POSITIONAL PROBABILITY

Highest probability from the activity?

Why?

POSITIONAL PROBABILITY This concept is also illustrated in

changes of state. Positional entropy increases from

solid to liquid to gas. Also applies to solution making.

Solutes and solvents in solution have more available volume to move around in. Thus the formation of solutions is favored.

PHASE CHANGE AT A CONSTANT TEMPERATUREΔSsurr = heat transferred = q temperature at which change occurs T

** where the heat supplied (endothermic) (q > 0) or evolved (exothermic) (q < 0) is divided by the temperature in Kelvins.** It is important here to note if the reaction is endothermic or exothermic.

PHASE CHANGE AT A CONSTANT TEMPERATURE Taking favored conditions into

consideration, the equation above rearranges into:

ΔSsurr = - ΔH T

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

PRACTICE FOURIn the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore. For example, iron is used to reduce antimony in sulfide ores:

Sb2S3(s) +3Fe(s) → 2Sb(s) +3FeS(s) ΔH = -125kJCarbon is used as the reducing agent for oxide ores:

Sb4O6(s) +6C(s) →4Sb(s) +6CO(g) ΔH = 778kJCalculate ΔSsurr for each of these reactions at 25°C and 1 atm.

ENTROPY SUMMARY

ΔS = + MORE DISORDER (FAVORED CONDITION)ΔS = - MORE ORDER Whether a reaction will occur spontaneously may

be determined by looking at the ΔS of the universe.

ΔSsystem + ΔSsurroundings = ΔSuniverse

IF ΔSuniverse is +, then reaction is spontaneous.IF ΔSuniverse is -, then reaction is NONspontaneous. 

EXAMPLE OF THISConsider 2 H2(g) + O2(g) → 2 H2O(g) Ignite this and the reaction is fast!ΔSsystem = -88.9J/K entropy declines (due mainly to 3→2 moles

of gas!). . . to confirm we need to know entropy of surroundings.

  ΔSsurr = q surr

T this comes from ΔH calculation: Hsystem = - 483.6 kJ

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

-ΔHsystem = ΔHsurr OR -(-483.6 kJ) = +483.6 kJ ΔS°surr = ΔH°surr = + 483.6 kJ = 1620 J/K

T 298 KNow we can find ΔS°universe

ΔSsystem + ΔSsurr = ΔSuniverse

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

EXAMPLE OF THIS

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

Bottom line: A process is spontaneous in spite of a negative entropy change as long as it is extremely exothermic. Sufficient exothermicity offsets system ordering.

GIBBS FREE ENERGY

Calculation of Gibb’s free energy is what ultimately decides whether a reaction is spontaneous or not.

NEGATIVE ΔG’s are spontaneous.

STANDARD FREE ENERGY CHANGE

G° is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states.

G° cannot be measured directly. The more negative the value for G°, the

farther to the right the reaction will proceed in order to achieve equilibrium.

Equilibrium is the lowest possible free energy position for a reaction.

METHOD # 1ΔG°rxn = ΣΔG°(products) - ΣΔG°(reactants)

This works the same way as enthalpy and entropy from tables of standard values. Standard molar free energy of formation. ΔG°f = 0 for elements in standard state.

METHOD # 2ΔG = ΔH - TΔS If G is -, the rxn is spontaneous in the

forward direction. If G = 0, the rxn is at equilibrium. If G is +, the rxn is nonspontaneous in the

forward direction, but the reverse rxn will be spontaneous.

G°f for elements at standard state (pure elements at 25ºC and 1 atm are assigned a value of zero).

METHOD # 3Hess’s law summation Works same as Hess’s in the enthalpy

section - sum up equations using the guidelines as before.

METHOD # 4G = ΔG° + RT ln(Q) Define terms: ΔG = free energy not at standard conditions ΔG° = free energy at standard conditions T = temperature in Kelvin R = universal gas constant 8.3145 J/mol∙K ln = natural log Q = reaction quotient: (for gases this is the partial

pressures of the products divided by the partial pressures of the reactants—all raised to the power of their coefficients) Q = [products]÷ [reactants].

METHOD # 5

“RatLink”: ΔG°= -RTlnK Terms: basically the same as above.

Here the system is at equilibrium, so ΔG = 0 and K represents the EQ constant under standard conditions. K = [products]÷ [reactants] still raised to power of coefficients.

METHOD #6“nFe”: ΔG°= - nFE° remember this one Terms: ΔG° = just like above—standard free

energy n = number of moles of electrons transferred

(look at ½ reactions) F = Faraday’s constant 96,485 Coulombs/mole

electrons E° = standard voltage ** one volt =

joule/coulomb**

PRACTICE FIVEGiven the following information, find the free energy of formation for the oxidation of water to produce hydrogen peroxide: 2 H2O(l) + O2(g) → 2 H2O2(l)

ΔG°f

H2O(l) -56.7 kcal/molO2(g) 0 kcal/molH2O2(l) -27.2 kcal/mol

ΔG°rxn = ΣΔG°(products) - ΣΔG°(reactants)

PRACTICE SIXConsider the reaction 2SO2(g) +O2(g)→2S)3(g) carried out at 25°C and 1 atm. Calculate ΔH°, ΔS°, and ΔG° using the following data:  Substance ΔH°f S° (J/K mol)SO2(g) -297 248SO3(g) -396 257O2(g) 0 205

ΔG = ΔH - TΔS

PRACTICE SEVENUsing the following data (at 25°C)Cdiamond(s) + O2(g) → CO2(g) ΔG°= -397 kJ Cgraphite(s) + O2(g) → CO2(g) ΔG°= -394 kJ Calculate ΔG° for the reaction: Cdiamond(s) →Cgraphite(s)

Hess’s law of summation

PRACTICE EIGHTOne method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases: CO(g) + 2H2(g) →CH3OH(l) Calculate ΔG at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol.

ΔG°f kJCH3OH -166H2 0CO -137

ΔG = ΔG° + RT ln (Q)

PRACTICE NINEThe overall reaction for the corrosion (rusting) of iron by oxygen is 4 Fe (s) + 3 O2 (g) ↔ 2 Fe2O3 (s). Using the following data, calculate the equilibrium constant for this reaction at 25°C.Substance ΔH°f S°(J/K mol)Fe2O3(s) -826 90Fe(s) 0 27O2(g) 0 205

ΔG° = -RTlnK

ΔG° = ΔH - T ΔS

Gibb’s equation can also be used to calculate the phase change temperature of a substance.

During the phase change, there is an equilibrium between phases so the value of ΔG° is zero.

PRACTICE TENFind the thermodynamic boiling point of H2O(l) → H2O(g) given the following information:Hvap = +44kJ Svap = 118.8J/K

ΔG° = ΔH - T ΔS

H, S, G AND SPONTANEITYG = H - TS

H is enthalpy, T is Kelvin temperature

Value of H

Value of TS

Value of G

Spontaneity

Negative Positive Negative

Spontaneous

Positive Negative Positive NonspontaneousNegative Negative ??? Spontaneous if the

absolute value of H is greater than the absolute value of TS (low temperature)

Positive Positive ??? Spontaneous if the absolute value of TS is greater than the absolute value of H (high temperature)

SUMMARY OF FREE ENERGY

ΔH ΔS Enegative positive Spontaneous at all temperaturespositive positive Spontaneous at high

temperaturesnegative negative Spontaneous at low

temperaturespositive negative Not spontaneous, ever

ΔG = + NOT SPONTANEOUSΔG = - SPONTANEOUS

Conditions of ΔG:

RELATIONSHIP OF ΔG TO K AND E

Equilibrium point occurs at the lowest value of free energy available to the reaction system

At equilibrium, G = 0 and Q = K

ΔG K E0 At EQ; K = 1 0

negative >1; products favored

positive

positive <1; reactants favored

negative

RELATIONSHIP OF ΔG TO K AND E Minimum free energy is reached when 75% of A

has been changed to B. Each point on the curve corresponds to the total

free energy of the system for a given combination of A and B.

FREE ENERGY AND WORK The maximum possible useful work

obtainable from a process at constant temperature and pressure is equal to the change in free energy. wmax = ΔG

The amount of work obtained is always less than the maximum.

Henry Bent's First Two Laws of Thermodynamics:

First law - You can't win, you can only break even

Second law - You can't break even.

FREE ENERGY AND WORK When the reaction is spontaneous, ΔG is the

energy available to do work such as moving a piston or flowing of electrons.

When ΔG is positive, and thus non-spontaneous, it represents the amount of work needed to make the process occur.

FREE ENERGY AND PRESSURE

Enthalpy, H, is NOT pressure dependent

Entropy, S is pressure dependent entropy depends on volume, so it

also depends on pressureSlarge volume > Ssmall volume

Slow pressure > Shigh pressure

TEMPERATURE DEPENDENCE OF K

000 )ln( STHKRTG

RS

TRH

RS

RTHK

0000 1)ln(

So, ln(K) 1/T