Entropy and Free Energy (Kotz Ch 20) - Lecture #2

10 Nov 97 Entropy & Free Energy (Ch 20) -

lect. 21

Entropy and Free Energy (Kotz Ch 20) - Lecture #2

• Spontaneous vs. non-spontaneous• thermodynamics vs. kinetics• entropy = randomness (So)

• Gibbs free energy (Go)• Go for reactions - predicting spontaneous direction

• thermodynamics of coupled reactions• Grxn versus Go

rxn

• predicting equilibrium constants from Gorxn


lect. 22

Entropy and Free Energy ( Kotz Ch 20 )

• How can we predict if a reaction can occur, given enough time?

• Note: Thermodynamics DOES NOT say how quickly (or slowly) a reaction will occur.

• To predict if a reaction can occur at a reasonable rate, one needs to consider:

• some processes are spontaneous; others never occur. WHY ?

THERMODYNAMICS

KINETICS

9-paper.mov20m02vd1.mov


lect. 23

Product-Favored Reactions

E.g. thermite reaction

Fe2O3(s) + 2 Al(s)

2 Fe(s) + Al2O3(s)

H = - 848 kJ

In general, product-favored reactions are exothermic.


lect. 24

Non-exothermic spontaneous reactions

But many spontaneous reactions or processes are endothermic . . .

NH4NO3(s) + heat NH4+ (aq) + NO3

- (aq)Hsol = +25.7 kJ/mol

or have H = 0 . . .


lect. 25

PROBABILITY - predictor of most stable state

WHY DO PROCESSES with H = 0 occur ?

Consider expansion of gases to equal pressure:

This is spontaneous because the final state,with equal # molecules in each flask, is much more probable than the initial state,with all molecules in flask 1, none in flask 2

SYSTEM CHANGES to state of HIGHER PROBABILITYFor entropy-driven reactions - the more RANDOM state.


lect. 26

Standard Entropies, So

• Every substance at a given temperature and in a specific phase has a well-defined Entropy• At 298o the entropy of a substance is called

So - with UNITS of J.K-1.mol-1

• The larger the value of So, the greater the degree of disorder or randomness

e.g. So (in J.K-1mol-1) : Br2 (liq) = 152.2

Br2 (gas) = 245.5

For any process: So = So(final) - So(initial)

So(vap., Br2) = (245.5-152.2) = 93.3 J.K-1mol-1


lect. 27

S (gases) > S (liquids) > S (solids)

So (J/K•mol)

H2O(gas) 188.8

H2O(liq) 69.9

H2O (s) 47.9

Ice Water

Vapour

Entropy and Phase


lect. 28

The entropy of a substance increases with temperature.

Molecular motions of heptane at different temps.

Entropy and Temperature

9_heptane.mov20m04an2.mov

Higher T means :• more randomness• larger S


lect. 29

Entropy and complexity

Increase in molecular complexity generally leads to increase in S.

9_alkmot.mov20m04an3.mov

SSoo (J/K•mol) (J/K•mol)

CHCH44 248.2248.2

CC22HH66 336.1 336.1

CC33HH88 419.4419.4


lect. 210

• Ionic Solids : Entropy depends on extent of motion of ions. This depends on the strength of coulombic attraction.

Entropy of Ionic Substances

• Entropy increases when a pure liquid or solid dissolves in a solvent.

NH4NO3(s) NH4+ (aq) + NO3

- (aq)

Ssol =

ion pairsion pairs SSoo (J/K•mol) (J/K•mol)

MgOMgO MgMg2+2+ / O / O2-2- 26.926.9

NaFNaF NaNa++ / F / F-- 51.551.5

So(aq. ions) - So(s) = 259.8 - 151.1= 108.7 J.K-1mol-1


lect. 211

Entropy Change in a Phase Changes

For a phase change,

S = q/Twhere q = heat transferred in phase change

H2O (liq) H2O(g)

For vaporization of water:

H = q = +40,700 J/mol

S = qT

= 40, 700 J/mol

373.15 K = + 109 J/K • mol


lect. 212

Consider 2 H2(g) + O2(g) 2 H2O(liq)

So = 2 So (H2O) - [2 So (H2) + So (O2)]

So = 2 mol (69.9 J/K•mol) - [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)]

So = -326.9 J/K

Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid.

Calculating S for a Reaction

So = So (products) - So (reactants)

If S DECREASES, why is this a SPONTANEOUS REACTION??


lect. 213

2nd Law of Thermodynamics

Suniverse = Ssystem + Ssurroundings

Suniverse > 0 for product-favored process

First calc. entropy created by matter dispersal (Ssystem)

Next, calc. entropy created by energy dispersal (Ssurround)

A reaction is spontaneous (product-favored) if S for the universe is positive.


lect. 214

Calculating S(universe)

2 H2(g) + O2(g) 2 H2O(liq)

Sosystem = -326.9 J/K

T

H- =

T

q =

systemgssurroundingssurroundin

ooS

K 298.15

J/kJ) kJ)(1000 (-571.7 - = gssurroundin

oS

Sosurroundings = +1917 J/K

Can calculate that Horxn = Ho

system = -571.7 kJ


lect. 215

Calculating S(universe) (2)

2 H2(g) + O2(g) 2 H2O(liq)

Sosystem = -326.9 J/K (less matter dispersal)

Sosurroundings = +1917 J/K (more energy dispersal)

The entropy of the universe increases so the reaction is spontaneous.

Souniverse = +1590 J/K


lect. 216

E = q + w

The Laws of Thermodynamics

0. Two bodies in thermal equilibrium are at same T

1. Energy can never be created or destroyed.

2. The total entropy of the UNIVERSE ( = system plus surroundings) MUST INCREASE in every spontaneous process.

STOTAL = Ssystem + Ssurroundings > 0

3. The entropy (S) of a pure, perfectly crystalline compound at T = 0 K is ZERO. (no disorder)

ST=0 = 0 (perfect xll)


lect. 217

Gibbs Free Energy, G

Suniv = Ssurr + Ssys

Suniv = Hsys

T + Ssys

Go = Ho - TSo

Multiply through by -T-TSuniv = Hsys - TSsys

-TSuniv = change in Gibbs free energy

for the system = Gsystem

Under standard conditions —The GibbsEquation


lect. 218

Standard Gibbs Free Energies, Gof

• Every substance in a specific state has a Gibbs Free Energy, G = H - TS• recall: only H can be measured. Therefore: there is no absolute scale for G• only G values can be determined

• Gof the Gibbs Free Energy of formation (from

elements) is used as the “standard value”

• We set the scale of G to be consistent with that

for H - Gof for elements in standard states = 0


lect. 219

Go < 0 for all SPONTANEOUS processes

Sign of G for Spontaneous processes

STOTAL = Ssystem + Ssurroundings > 0

2nd LAW requirement for SPONTANEITY is :

Multiply by T TSsystem + TSsurroundings > 0

and Ssurroundings = Hosystem/T

Thus TSsystem - Hosystem > 0

Multiply by -1 (->reverse > to <), drop subscript “system”

Ho -TS < 0 and Go = Ho

-TS


lect. 220

Go = Ho - TSo

• change in Gibbs free energy =

(total free energy change for system - free energy lost in disordering the system)

• If reaction is exothermic (Ho is -ve) and entropy increases (So is +ve), then

Go must be -ve and reaction CAN proceed.• If reaction is endothermic (Ho is +ve), and entropy decreases (So is -ve), then Go must be +ve; reaction CANNOT proceed.

Sign of Gibbs Free Energy, G


lect. 221

Gibbs Free Energy changes for reactions

Ho So Go Reaction

exo (-) increase(+) - Product-favored

endo(+) decrease(-) + Reactant-favored

exo (-) decrease(-) ? T dependent

endo(+) increase(+) ? T dependent

Spontaneous in last 2 cases only ifTemperature is such that Go < 0

Go = Ho - TSo


lect. 222

Methods of calculating G

Two methods of calculating Go

GGoorxnrxn = = GGff

oo (products) - (products) - G Gffoo (reactants) (reactants)

a) Determine Horxn and So

rxn and use Gibbs

equation.

b) Use tabulated values of free energies of

formation, Gfo.

Go = Ho - TSo


lect. 223

Calculating Gorxn

EXAMPLE: Combustion of acetylene

C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(g)

From standard enthalpies of formation: Horxn = -1238 kJ

From standard molar entropies: Sorxn = - 0.0974 kJ/K

Calculate Gorxnfrom Go = Ho - TSo

Gorxn = -1238 kJ - (298 K)(-0.0974 kJ/K)

= -1209 kJ

Reaction is product-favored in spite of negative So

rxn. Reaction is “enthalpy driven”


lect. 224

Is the dissolution of ammonium nitrate product-favored?

If so, is it enthalpy- or entropy-driven?

EXAMPLE 2:

NH4NO3(s) NH4NO3(aq)

Calculating Gorxn for NH4NO3(s)

9_amnit.mov20 m07vd1.mov


lect. 225

Gorxn for NH4NO3(s) NH4NO3(aq)

From tables of thermodynamic data we find

Horxn = +25.7 kJ

Sorxn = +108.7 J/K or +0.1087 kJ/K

Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)

= -6.7 kJ

Reaction is product-favored

. . . in spite of positive Horxn.

Reaction is “entropy driven”


lect. 226

Calculating Gorxn

EXAMPLE 3: Combustion of carbon

C(graphite) + O2(g) CO2(g)

Gorxn = Gf

o(CO2) - [Gfo(graph) + Gf

o(O2)]

Gorxn = -394.4 kJ - [ 0 + 0]

Note that free energy of formation of an element in its standard state is 0.

Gorxn = -394.4 kJ

Reaction is product-favored as expected.

GGoorxnrxn = = GGff

oo (products) - (products) - GGffoo (reactants) (reactants)


lect. 227

Free Energy and Temperature

2 Fe2O3(s) + 3 C(s) 4 Fe(s) + 3 CO2(g)

Horxn = +467.9 kJ So

rxn = +560.3 J/K

Gorxn = 467.9 kJ - (298K)(0.560kJ/K) = +300.8 kJ

Reaction is reactant-favored at 298 K

At what T does Gorxn just change from (+) to (-)?

i.e. what is T for Gorxn = 0 = Ho

rxn - TSorxn

If Gorxn = 0 then Ho

rxn = TSorxn

so T = Ho/So ~ 468kJ/0.56kJ/K = 836 K or 563oC


lect. 228

Go for COUPLED CHEMICAL REACTIONS

Reduction of iron oxide by CO is an example ofusing TWO reactions coupled to each other in orderto drive a thermodynamically forbidden reaction:

Fe2O3(s) 4 Fe(s) + 3/2 O2(g) Gorxn = +742 kJ

3/2 C(s) + 3/2 O2 (g) 3/2 CO2(g) Gorxn = -592 kJ

with a thermodynamically allowed reaction:

Overall : Fe2O3(s) + 3/2 C(s) 2 Fe(s) + 3/2 CO2(g)

Gorxn= +301 kJ @ 25oC

BUT Gorxn < 0 kJ for T > 563oC

See Kotz, pp933-935 for analysis of the thermite reaction


lect. 229

Other examples of coupled reactions:

Copper smelting

Cu2S (s) 2 Cu (s) + S (s) Gorxn= +86.2 kJ

(FORBIDDEN) Couple this with:S (s) + O2 (g) SO2 (s) Go

rxn= -300.1 kJ

Overall: Cu2S (s) + O2 (g) 2 Cu (s) + SO2 (s)

Gorxn= +86.2 kJ + -300.1 kJ = -213.9 kJ (ALLOWED)

Coupled reactions VERY COMMON in Biochemistry :e.g. all bio-synthesis driven by

ATP ADP for which Horxn = -20 kJ

Sorxn = +34 J/K

Gorxn = -30 kJ @ 37oC


lect. 230

Thermodynamics and Keq

• Keq is related to reaction favorability.

• If Gorxn < 0, reaction is product-favored.

• Gorxn is the change in free energy as reactants

convert completely to products.

• But systems often reach a state of equilibrium in which reactants have not converted completely to products.

• How to describe thermodynamically ?


lect. 231

Grxn versus Gorxn

Under any condition of a reacting system, we can define Grxn in terms of the REACTION QUOTIENT, Q

Grxn = Gorxn + RT ln Q

At equilibrium, Grxn = 0. Also, Q = K. Thus

If Grxn < 0 then reaction proceeds to rightIf Grxn > 0 then reaction proceeds to left

Gorxn = - RT lnK


lect. 232

2 NO2 N2O4

Gorxn = -4.8 kJ

• pure NO2 has Grxn < 0.

• Reaction proceeds until Grxn = 0 - the minimum in G(reaction) - see graph.

• At this point, both N2O4 and NO2 are present, with more N2O4.

• This is a product-favored reaction.

Thermodynamics and Keq (2)

9_G_NO2.mov20m09an1.mov


lect. 233

N2O4 2 NO2

Gorxn = +4.8 kJ

• pure N2O4 has Grxn < 0.

• Reaction proceeds until Grxn

= 0 - the minimum in G(reaction) - see graph.

• At this point, both N2O4 and NO2 are present, with more NO2.

• This is a reactant-favored reaction.


9_G_N2O4.mov20m09an2.mov


lect. 234


Keq is related to reaction favorability and so to Go

rxn.

The larger the value of Gorxn the larger the

value of K.

Gorxn = - RT lnK

where R = 8.31 J/K•mol


lect. 235

Calculate K for the reaction

N2O4 2 NO2 Gorxn = +4.8 kJ

Gorxn = +4800 J = - (8.31 J/K)(298 K) ln K

Gorxn = - RT lnK

lnK = -4800 J

(8.31 J/K)(298K) = - 1.94


When Gorxn > 0, then K < 1 - reactant favoured

When Gorxn < 0, then K >1 - product favoured

K = 0.14


lect. 236

Entropy and Free Energy (Kotz Ch 20)

• Spontaneous vs. non-spontaneous• thermodynamics vs. kinetics• entropy = randomness (So)

• Gibbs free energy (Go)• Go for reactions - predicting spontaneous direction

• thermodynamics of coupled reactions• Grxn versus Go

rxn

• predicting equilibrium constants from Gorxn

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Entropy and Free Energy (Kotz Ch 20) - Lecture #2

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