Free Energy and Thermodynamics Perpetual motion is forbidden by the Laws of Thermodynamics 1

Free Energy and Thermodynamics

Perpetual motion is forbidden by the Laws of Thermodynamics1

Chapter 16

Spontaneity, Entropy And Free Energy1st Law of Thermodynamics

Energy can be neither created nor destroyed.Energy of the universe is constant

However not all energy is usable

The 1st Law is an energy bookkeeping systemHelps answer questions:

How much energy in involved?Does energy flow in or out of system?What form does the energy assume?

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Spontaneous• A reaction is spontaneous if it occurs without outside

intervention Ex. Fe + H2O + O2 Fe2O3 (Rust)

• Does not mean fast• We will be using Thermochemistry (Ch.6), Kinetics

(Ch.12) & Thermodynamics (Ch. 16) to describe a reaction completely.

• Thermodynamics tells the direction not speed• Thermodynamics compares initial & final states• Kinetics describes pathway between reactant &

product. 3

Formula Review

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( Energy) ( ) ( )Internal heat workE q w

( )EnthalpyH E P V

q mc T [ ]

[ ][ ]

nARate k A

t

Spontaneity• A process that is spontaneous in one direction will

not be spontaneous in the other.• Ex. If wood burns into CO2 and H2O, Why can we not put CO2

and H2O together and make wood?

• Processes will proceed without outside assistance• If steel is exposed to air and moisture it rusts

• It’s independent of time• The characteristic common in all spontaneous

processes is an increase in the property called Entropy (S)

If S(universe) > 0, the reaction is spontaneous 5

Entropy• Entropy is a measure of molecular randomness

or disorder• High Disorder = High Entropy• Entropy is a thermodynamic function that

describes the number of arrangements (positions) available to a system, it’s probability.

• Entropy is a measure of chaos in a system• Entropy is a state function• Driving force in all spontaneous processes is

increase in Entropy (S) of the universe 6

Entropy• Formal definition: Entropy (S) is a thermodynamic

function that increases with the number of energetically equivalent ways to arrange the components of a system to achieve a particular state.

• First expressed by Ludwig Boltzmann in early 1900S = k ln W

• k is Boltzmann constant = gas constant (8.31) / Avogadro’s number = 1.38 X 10-23 J/K

• W = number of energetically equivalent ways7

• 2 possible arrangements• 50 % chance of finding

the left empty• Each configuration that

gives a particular arrangement is called a microstate

1 MOLECULE OF GAS

Probability Consider the number of arrangements for a

system in the following evacuated bulbs

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4 possible arrangements25% chance of finding

the left empty50 % chance of them

being evenly dispersed25% chance of finding

the right empty

2 MOLECULES OF GAS

See Table 16.1 p.753

Probability

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Positional Entropy

• Positional Entropy (S) – depends upon the number of microstates a set of molecules can have in space1. Solids have the fewest possible number of microstates

in space, atoms are very ordered, crystal structures2. Liquids have many more ways for molecules to be

arranged than a solid3. Gases have a huge number of positions possible

S solid < S liquid << S gas• A chemical system proceeds in a direction

that increases the entropy of the universe.10

Positional Entropy• Choose the substance with the higher

positional entropy:1. CO2(s) vs. CO2(g)

2. N2(g) at 1 atm vs. N2(g) at 1.0 X 10-2 atm

• As the change in entropy (∆S) becomes more positive the system/universe becomes more disorder.

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Entropy Values @ 298K J/mol K

Solids Liquids GasesC(graph.) 5.7 H2O 69.9 H2 130.7

Ca 41.4 CH3OH 126.8 O2 205.1

MgO 26.9 Br2 152.2 Cl2 223.0

NaCl 72.1

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“Entropy Ain’t What It Used To Be”• 2nd Law of Thermodynamics: Entropy of the

universe is always increasing• If process is spontaneous Total entropy must increase

• ∆S(universe) is always increasing

Entropy defined in terms of probability– Describes the number of arrangements available to a

system existing in a given state– The most likely of the particles is the most random– Nature proceeds toward the system with the highest

number of possible states

Universe system surroundingsS S S

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Predicting Entropy Changes• Predict the sign of the entropy ∆S for each of the

following processes.1. Solid sugar is added to water to form a solution.2. Iodine vapor condenses on a cold surface to form a crystal

Remember: S solid < S liquid << S gas

Use this formula to help determine ∆S

∆S = S(Final) – S(Initial)

Which has a greater number of arrangement:

Final state or Initial state14

Entropy• Solutions form because there are many more

possible arrangements of dissolved pieces than if they stay separate.

Remember 2nd Law of Thermodynamics, Entropy of universe is always increase

Suniverse = Ssystem + Ssurroundings

IfSuniv process is spontaneous in forward directionIfsuniv process is spontaneous in opposite directionIf suniv = 0, process has no tendency to occur = equilibrium

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Exothermic processes (heat out) Ssurr is positive

Endothermic processes (heat ) Ssurr is negative

• Consider this process: H2O(l)H2O(g)

• Is it spontaneous? i.e. Is suniv positive)?• How much volume will 1 mole of liquid and gas occupy? (Stoichiometry)

• Ssys is positive (liquid gas, entropy)

• Ssurr is negative (vaporization is an endothermic process)

ssys = sproducts - sreactants

suniv = ssys + ssurr

Effect of Temperature on Spontaneity

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Entropy Changes In The Surroundings• Energy changes in the surroundings are primarily

determined by flow of heat. i.e the sign of ∆Ssurr depends on direction of heat flow.– An exothermic process is favored because by giving up

heat, the entropy of the surroundings increases.– Summary

Quantity of Heat JExothermic Process S =+

Temperature Ksurr

H

T

Quantity of Heat JEndothermic Process S = -

Temperature Ksurr

H

T

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Entropy Changes In The Surroundings

• How much of an impact the transfer of heat is on the surroundings depends on how high the temperature is.

• If the transfer occurs at low temperature, Ssurr will be greater.

• Magnitude of Ssurr depends directly on quantity of heat transferred

• Magnitude of Ssurr depends inversely on temperature

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Enthalpy H• Enthalpy (H) is the internal energy (E) of a system

plus the pressure times volume (PV)• H=E+PV• H = a measure of the total change of internal

energy of a system.• H heat flow into system (endothermic)

• H heat flow into system (exothermic)

• Entropy (S) and Enthalpy (H): two factors that determine spontaneity

• Nature will always transfer thermodynamic energy from a higher to a lower temperature 19

Determining Ssurr

In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore @ 25oC

Iron is used to reduce antimony in sulfide ores. Sb2S3(s) + 3Fe(s) 2Sb(s) + 3FeS(s) H= -125kJ

Carbon is used as the reducing agent for oxide ores.Sb4O6(s) + 6C(s) 4Sb(s) + 6CO(g) H= 778kJ

Label RedOx Values & Calculate Ssurr for each reaction

surrH

ST

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Spontaneous or Not?Ssys Ssurr Suniv Spontaneous?

+ + + Yes

- - - Yes in Reverse

+ - ?Yes If

Ssys>Ssurr

- + ?Yes If

Ssys<Ssurr21

Gibb's Free Energy• Gibb's Free Energy is defined as: G=H-TS

G=Free Energy H=Enthalpy T=Temp.(K) S=Entropy

G=H-TS (@ constant temperature)

(H & S without subscripts means system)

• Study the derivation of formulas on p. 760

• If G < 0 @ constant T and P, the Process is spontaneous in which the free energy is decreasing

-G means +S

UNIVG

ST

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Let’s Check

• For the reaction H2O(s) H2O(l)• So = 22.1 J/K mol • Ho =6030 J/mol

• Calculate G at 10oC, 0oC, & -10oC (convert to K)Superscripto means substance is in standard state

• Use equation G0=H0-TS0

• Spontaneity can be predicted from the sign of H and S.

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At What Temperature Is The Following Process Spontaneous?

Br2(l) Br2(g)

∆H0 = 31.0 kJ/mol∆S0 = 93.0 J/K

What is the normal boiling point for Br2(l)

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At What Temperature Is The Following Process Spontaneous?

CH3OH(l) CH3OH(g)

1.Get Ho and So values from Appendix 42.Caution: Make sure your units are compatible3.Plug values into formula

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Summary G=H-TS

S H Spontaneous?

+ - At all temp.

+ + At high temp.“entropy driven”

- - At low temp.“enthalpy driven”

- + Not at any temp.Reverse at all temp.

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• Predict the sign of ∆S for each of the following reactions. 1. Thermal decomposition of solid calcium carbonate

CaCO3(s) CaO(s) + CO2(g)

2. The oxidation of SO2 in air

2SO2(g) + O2(g) → 2SO3(g)

In general, when a reaction involves gaseous molecules, the change in positional entropy is

dominated by the relative number of molecules of reactants and products.

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Third Law of ThermodynamicsEntropy (S) Of A Pure Crystal At 0K = 0

• Walther Hermann Nernst (1864-1941)• Gives us a starting point• All others must be >0• Standard Entropies values So (@298 K & 1 atm) of

substances are listed Appendix 4• Entropy is a state function of a system and ∆S of a

given reaction can be calculated using the formula:

. .reaction

o o op prod r reactS n S n S

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Calculate ∆So

• Predict sign of ∆So

• Calculate ∆So at 25oC2NiS(s) + 3O2(g) → 2SO2(g) + 2 NiO(s)

Substance So(J/Kmol)

SO2(g) 248

NiO(s) 38

O2(g) 205

NiS(s) 53 29

Calculate ∆So

• Calculate ∆So for the reduction of aluminum oxide by hydrogen gas

Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g)

Substance So(J/Kmol)

Al2O3(s) 51

H2(g) 131

Al(s) 28

H2O(g) 18930

Free Energy and Chemical Reactions• Go = standard free energy change.• Free energy change that will occur if reactants in

their standard state turn to products in their standard state.

• Go can’t be measured directly, can be calculated from other measurements.

• Go=Ho-TSo

• Can calculate Ho using enthalpy of formation f equation p.246

( ) (reactants)

o o o

f products r fpH n H n H

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Calculating Ho, So, & Go

• Consider the reaction:2SO2(g) + O2(g) 2SO3(g)

Carried out at 25oC & 1 atm, Use the following data:

Substance Hfo(kJ/mol) So(J/Kmol)

SO2 -297 248

SO3 -396 257

O2 0 20532

Hess’s Law “More than one way to skin a cat”

Because ∆G is a state function you can obtain ∆G for reactions using Hess’s Law

2CO(g) + O2(g) 2CO2(g)

Use a couple of related equation and manipulate them to get ∆Go

2CH4(g) + 3O2(g) 2CO(g) + 4H2O(g) ∆Go = -1088kJCH4(g) + 2O2(g) CO2(g) + 2H2O(g) ∆Go = -801kJ

See Ch 6.3 p. 242 for review on Hess’s Law

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Calculate Go

• Using the following data at 25oC

Cdiamond(s) + O2(g) CO2(g) Go = -397 kJ

Cgraphite(s) + O2(g) CO2(g) Go = -394 kJ

Calculate Go for the reaction:Cdiamond(s) Cgraphite(s)

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Free Energy in Reactions• Standard free energy of formation (Go

f) is change in free energy that accompanies the formation of 1 mol of that substance from reactants and products in their standard state Go

f

• The standard free energy of formation for any element in its standard state is 0, (see table on p.A19)

• Use the following formula:

. .o o o

p prod r reactG n G n G 35

Calculate Go

Methanol is a high-octane fuel used in high-performance racing engines.

Calculate Go for the reaction, given the Gfo below:

2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)

Substance

Gfo

CH3OH(g) -163

O2(g) 0

CO2(g) -394

H2O(g) -22936

Free Energy is Dependent on Pressure

• Entropy depends on pressure because it effects volume.

• Slarge volume > Ssmall volume

• Because volume and pressure are inversely related

• Slow pressure> Shigh pressure

Study the Derivation of G = Go +RTln(Q) 37

Calculate G

• One method for synthesizing methanol CH3OH(l)

involves reacting carbon monoxide and hydrogen gas @25oC. Calculate G for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to methanol.

CO(g) + H2(g) → CH3OH(l)

1st calc. Go Get Gfo values p. A19 & formula p. 769

2nd calc. G Using G = Go +RTln(Q)

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How far?

As a reaction proceeds toward Equilibrium, the G changes until it reaches zero which is the Equilibrium point.

G = Gprod. - Greact. = 0

At Equilibrium G = 0, Q = KFor reactions run under “nonstandard conditions”

G = Go + RTlnQR = 8.31J/molKT = Temp in K

c d

a b

C DQ

A B

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Free Energy & Equilibrium• At equilibrium, ∆G = 0 and k = Q so you can plug in 0 for ∆G

and k for Q, rearranging the formula you get new formula that lets you quantitatively relate free energy to equilibrium.

lnoG RT k Like wise, you can relate the pressure of reactants and products to the equilibrium constant k with the following equation:

. .

. .

( )( )

( )( )prod prod

react react

P Pk

P P

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Quantitative Relationship Between Free Energy And Equilibrium Constant

Go KGo = 0 K = 1Go < 0 K > 1Go > 0 K < 1

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Free Energy & Equilibrium• Consider the ammonia synthesis reaction

• Where Go = -33.3 kJ per mole of N2 consumed at 25oC. For each of the following mixtures of reactants and products at 25oC, predict the direction in which the system will shift to reach equilibrium.

1. PNH3 = 1.00 atm, PN2 = 1.47 atm, PH2 = 1.00x10-2 atm

2. PNH3=1.00 atm, PN2=1.00 atm, PH2= 1.00 atm

2( ) 2( ) 3( )3 2g g gN H NH

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Temperature Dependence Of K

Go = -RTln(K) = Ho - TSo

• Rearrangement gives

• A plot of ln(K) vs 1/T yields a straight line(Y=mx+b)

1ln

o oH SK

R T R

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Free energy And Work

• Free energy is energy free to do work• The maximum amount of work possible at a

given temperature and pressure is:wmax = G

• Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.

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Summary of Thermodynamic Laws

First Law: You can’t win, you can only break even

Second Law: You can’t break even

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