Chapter 17

Free Energy and

Thermodynamics

Goals

Entropy (S, S) and spontaneity

Free energy; G, Go

G, K, product- or reactant-favored

Review: H (Enthalpy) and the 1st Law of Thermodynamics

Chemical Equilibria (ch. 14, etc)

3

First Law of Thermodynamics

• First Law of Thermodynamics: Energy cannot be created or destroyedthe total energy of the universe cannot changeit can be transfered from one place to another

Euniverse = 0 = Esystem + surroundings

system = reactants & products

surroundings = everything else

(the transfer of energy from one to the other does not change the energy of the universe)

4

First Law of ThermodynamicsFor an exothermic reaction, heat from the system

goes into the surroundings• two ways energy can be “lost” from a system,

converted to heat, qused to do work, w Energy conservation requires that the energy change in the

system = heat exchanged + work done on the system. E = q + w (E = internal energy change) E = H – PV (at const. P, qp = H, enthalpy

change)• State functions (H, P, V). q and w are not.

internal energy change (E) independent of how done

5

Enthalpy, H• related to (includes) the internal energy• H generally kJ/mol• stronger bonds = more stable molecules• if products more stable than reactants, energy

released; exothermic H = negative

• if reactants more stable than products, energy absorbed; endothermic H = positive

• The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions.

• Hess’ Law: H°rxn = (Hf°prod) - (Hf°react)

6

Thermodynamics and Spontaneity• thermodynamics predicts whether a process will

proceed (occur) under the given conditionsspontaneous process

nonspontaneous process does not occur under specific conditions.

• spontaneity is determined by comparing the free energy (G) of the system before the reaction with the free energy of the system after reaction. if the system after reaction has less free energy than

before the reaction, the reaction is thermodynamically favorable.

• spontaneity ≠ fast or slow (rate); this is kinetics

7

Spontaneous Nonspontaneousice melts @ 25oC water freezes @ 25oC

2Na(s) + 2H2O(l) H2(g) + 2NaOH(aq) H2(g) + 2NaOH(aq) 2Na(s) + 2H2O(l)

ball rolls downhill ball rolls uphill

water freezes @ -10 oC ice melts @ -10oC

8

Diamond → Graphite

Graphite is thermodynamically more stable than diamond, so the conversion of diamond into

graphite is spontaneous – but it’s kinetically too slow (inert) it will never happen in many, many

generations.

kinetics

Spontaneity:direction & extent

kinetics:how fast

9

Factors Affecting Whether a Reaction Is Spontaneous

• The two factors that determine the thermodynamic favorability are the enthalpy and the entropy.

• The enthalpy is a comparison of the bond energy of the reactants to the products. bond energy = amount needed to break a bond. H

• The entropy factors relate to the randomness/orderliness of a system S

• The enthalpy factor is generally more important than the entropy factor

Substance Hf° kJ/mol

Substance Hf° kJ/mol

Al(s) 0 Al2O3 -1669.8 Br2(l) 0 Br2(g) +30.71

C(diamond) +1.88 C(graphite) 0 CO(g) -110.5 CO2(g) -393.5 Ca(s) 0 CaO(s) -635.5 Cu(s) 0 CuO(s) -156.1 Fe(s) 0 Fe2O3(s) -822.16 H2(g) 0 H2O2(l) -187.8

H2O(g) -241.82 H2O(l) -285.83 HF(g) -268.61 HCl(g) -92.30 HBr(g) -36.23 HI(g) +25.94

I2(s) 0 I2(g) +62.25 N2(g) 0 NH3(g) -46.19 NO(g) +90.37 NO2(g) +33.84 Na(s) 0 O2(g) 0 S(s) 0 SO2(g) -296.9

11

Entropy, S

• Entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases. S generally in J/K (joules/K)

• S = k lnWk = Boltzmann Constant (R/NA) = 1.38 10-23 J/K

W is the number of energetically equivalent ways, (microstates). It is unitless.

Entropy is usually described as a measure of the randomness or disorder; the greater the disorder of a system, the greater its S.The greater the order the smaller its S.

12

Entropy & Microstates, WEnergetically Equivalent States for the Expansion of a Gas (4 gas molecules) 1 microstate

1 microstate

6 microstates(most probable distribution)

S = k ln W

S = k ln Wf - k ln Wi

if Wf > Wi , S > 0 & entropy increases.

i

f

W

Wk lnS

13

Changes in Entropy, S• entropy change is favorable when the result is a

more random system (State C: higher entropy). S is positive (S > 0)

Some changes that increase the entropy are:rxns where products are in a more disordered state.

(solid > liquid > gas) less order(solid< liquid < gas) larger S (disorder)

reactions which have larger numbers of product molecules than reactant molecules.

increase in temperature (more movement)solids dissociating into ions upon dissolving

14

Changes in Entropy in a System(melting)

Particles fixed in space

Particles can occupy many positions

15

Changes in Entropy in a System(vaporization)

Particles occupy more space (larger volume)

16

Changes in Entropy in a System (solution process)

Structure of solute and solvent disrupted (also more solute particles)

17

Predict entropy change for a process/reaction

For which process/reaction is S negative?

Freezing ethanolMixing CCl4 with C6H6

Condensing bromine vapor

2O3(g) 3O2(g)

4Fe(s) + 3O2(g) 2Fe2O3(s)

2H2O2(aq) 2H2O(l) + O2(g)

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

2NH3(g) N2(g) + 3H2(g)

entropy dec

entropy dec

entropy dec

18

The 2nd Law of Thermodynamics

• The entropy of the universe increases in a spontaneous process. Suniverse = Ssystem + Ssurroundings > 0

Suniverse = Ssystem + Ssurroundings = 0 (equilibrium)

If Ssystem >> 0, Ssurroundings < 0 for Suniverse > 0! If Ssystem < 0, Ssurroundings >> 0 for Suniverse > 0!

• the increase in Ssurroundings often comes from the heat released in an exothermic reaction, system < 0.

19

The 3rd Law of Thermodynamics

S = k ln W

S = k ln W = k ln 1 = 0

S = Sf – Si; where Si = 0 @ 0 K

the absolute entropy of a substance is always (+) positive at the new T

-allows determination of entropy of substances.(W = 1, there is only one way to arrange the particles to form a perfect crystal)

the 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K

20

Standard Entropies

• S°

• entropies for 1 mole at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution

Values can be used to calculate the standard

entropy change for a reaction, Sorxn (= So

sys)

Substance S° J/mol-K

Substance S° J/mol-K

Al(s) 28.3 Al2O3(s) 51.00 Br2(l) 152.3 Br2(g) 245.3

C(diamond) 2.43 C(graphite) 5.69 CO(g) 197.9 CO2(g) 213.6 Ca(s) 41.4 CaO(s) 39.75 Cu(s) 33.30 CuO(s) 42.59 Fe(s) 27.15 Fe2O3(s) 89.96 H2(g) 130.58 H2O2(l) 109.6

H2O(g) 188.83 H2O(l) 69.91 HF(g) 173.51 HCl(g) 186.69

HBr(g) 198.49 HI(g) 206.3 I2(s) 116.73 I2(g) 260.57 N2(g) 191.50 NH3(g) 192.5 NO(g) 210.62 NO2(g) 240.45 Na(s) 51.45 O2(g) 205.0 S(s) 31.88 SO2(g) 248.5

22

Trends: Standard EntropiesMolar Mass

• For monatomic species, the larger the molar mass, the larger the entropy

• available energy states more closely spaced, allowing more dispersal of energy through the states

23

Trends: Standard EntropiesStates

• the standard entropy of a substance in the gas phase is greater than the standard entropy of the same substance in the solid or liquid phase at a particular temperature

SubstanceS°,

(J/mol∙K)

H2O (l) 70.0

H2O (g) 188.8

24

Trends: Standard EntropiesAllotropes

• the more highly ordered form has the smaller entropy

-different forms of an element

25

Trends: Standard EntropiesMolecular Complexity (inc # of atoms)

larger, more complex molecules generally have larger entropy

•more available energy states, allowing more dispersal of energy through the states

SubstanceMolar

MassS°,

(J/mol∙K)

Ar (g) 39.948 154.8

NO (g) 30.006 210.8

26

Trends: Standard EntropiesDissolution

• dissolved solids generally have larger entropy

• distributing particles throughout the mixture

SubstanceS°,

(J/mol∙K)

KClO3(s) 143.1

KClO3(aq) 265.7

27

Q. Arrange the following in order of increasing entropy @ 25oC! (lowest to highest)

Ne(g), SO2(g), Na(s), NaCl(s) and H2(g)

Na(s) < NaCl(s) < H2(g) < Ne(g) < SO2(g)

Q . Which has the larger entropy in each pair?a) Li(s) or Li(l)

b) C2H5OH(l) or CH3OCH3 (l)

c) Ar(g) or Xe(g)

d) O2(g) or O3(g)

Calculate S for the reaction4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

S is +, as you would expect for a reaction with more gas product molecules than reactant molecules

standard entropies from Appendix IIB

S, J/K

Check:

Solution:

Concept Plan:

Relationships:

Given:Find:

SSNH3, SO2, SNO, SH2O,

reactantsproducts SSS rp nn

K

J

K

J

K

J

K

J

K

J

)(O)(NH)O(H)NO(

reactantsproducts

8.178

)]2.205(5)8.192(4[)]8.188(6) 8.210(4[

)]S(5)S(4[)]S(6)S(4[

SSS

232

gggg

rp nn

Substance S, J/molK

NH3(g) 192.8

O2(g) 205.2

NO(g) 210.8

H2O(g) 188.8

Consider 2 HConsider 2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(liq) @ 25O(liq) @ 25ooCC

∆∆SSoo = 2 S = 2 Soo (H (H22O) - [2 SO) - [2 Soo (H (H22) + S) + Soo (O (O22)])]

∆∆SSoo = 2 mol (69.9 J/K•mol) - = 2 mol (69.9 J/K•mol) - [2 mol (130.6 J/K•mol) + [2 mol (130.6 J/K•mol) +

1 mol (205.0 J/K•mol)]1 mol (205.0 J/K•mol)]

∆So = -326.4 J/KNote that there is a Note that there is a decrease in S decrease in S because 3 mol of because 3 mol of

gas give 2 mol of liquid.gas give 2 mol of liquid.

Calculating ∆SCalculating ∆Soo for a Reaction for a Reaction

∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)

The more negative Hsyst and the lower the temperature the higher

(more positive) Ssurr 30

Temperature Dependence of Ssurroundings

system < 0 (exothermic), it adds heat to the surroundings, increasing the entropy of the surroundings (Ssurroundings > 0 )

system > 0 (endothermic), it takes heat from the surroundings, decreasing the entropy of the surroundings (Ssurroundings < 0 )

T

ΔHΔS system

gssurroundin

systemgssurroundin ΔHΔS

T

1gssurroundinΔS

2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(liq) @ 25 O(liq) @ 25 ooCC

∆Sosystem = -326.4 J/K

K 298

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

oSK 298

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

oS

Calculating ∆SCalculating ∆Soo for the surroundings for the surroundings

∆∆SSoosurroundingssurroundings = +1917 J/K = +1917 J/K

Can calculate ∆HCan calculate ∆Hoosystemsystem = ∆H = ∆Hoo

rxnrxn = -571.7 kJ = -571.7 kJ (also from tabulated data)(also from tabulated data)

T

ΔHΔS system

gssurroundin

2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(liq) @ 25O(liq) @ 25ooCC

∆∆SSoosystemsystem = -326.4 J/K = -326.4 J/K

∆∆SSoosurroundingssurroundings = +1917 J/K = +1917 J/K

∆∆SSoouniverse universe = +1591 J/K= +1591 J/K

The entropy of theThe entropy of the

universe is increasing,universe is increasing,

so the reaction is so the reaction is

spontaneous ( product-spontaneous ( product-

favored)favored). . (see slide # 18)(see slide # 18)

Suniverse = Ssystem + Ssurroundings

Given Given SSoosurrsurr , , SSoo

syssys and T, and T,

determine determine SSoouniv univ and predict and predict

if the reaction will be if the reaction will be spontaneous.spontaneous.

K

J3K

kJsurr

syssurr

106.86 86.6S

K 298

kJ 2044

T

HS

The reaction C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) has Hrxn = -2044 kJ at 25°C.

Calculate the entropy change of the surroundings.

combustion is largely exothermic, so the entropy of the surroundings should increase (inc in # gas mol)

Hsystem = -2044 kJ, T = 298 K

Ssurroundings, J/K

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

ST, H

T

ΔHΔS sys

surr

Spontaneous or Not?

originally: originally: Suniverse = Ssystem + Ssurroundings

but but

originally: originally: Suniverse = Ssystem + Ssurroundings

but but

Suniverse = Ssystem – system/T

system Ssystem Spontaneous?ExothermicHsys < 0

Less orderSsys > 0

Spontaneous under all conditions; Suniv > 0

ExothermicHsys < 0

More orderSsys < 0

Favorable at low T

EndothermicHsys > 0

Less orderSsys > 0

Favorable at high T

EndothermicHsys > 0

More orderSsys < 0

Not spontaneous under any conditions Suniv < 0

T

ΔHΔS sys

surr

35

Without doing any calculations, determine the sign of Ssys and Ssurr for each reaction. Predict under what temperatures (all T, low T, or high T) the reaction will be spontaneous.

2CO(g) + O2(g) 2CO2(g) rxn = -566.0 kJ

2NO2(g) O2(g) + 2NO(g) rxn = +113.1 kJ

Suniverse = Ssystem + (--sys/T) /T) Suniverse = Ssystem + (--sys/T) /T)

Ssystem = (-); 3 mol gas form 2 mol gas

Ssurr = (+); spontaneous @ low T

Ssystem = (+); 2 mol gas form 3 mol gas

Ssurr = (-); spontaneous @ high T

Ssurr = (--sys/T) /T)

36

Without doing any calculations, determine the sign of Ssys and Ssurr for each reaction. Predict under what temperatures (all T, low T, or high T) the reaction will be spontaneous.

2H2(g) + O2(g) 2H2O(g) rxn = -483.6 kJ

CO2(g) C(s) + O2(g) rxn = +393.5 kJ

Suniverse = Ssystem + (--sys/T) /T) Suniverse = Ssystem + (--sys/T) /T)

Ssystem = (-); 3 mol gas form 2 mol gasSsurr = (+) ; spontaneous @ low T

Ssystem = (-); complicated gas forms a solid & gas

Ssurr = (-); nonspontaneous @ all T

Ssurr = (--sys/T) /T)

37

At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings, if Ho

rxn = -127 kJ and Sorxn = 314 J/K.

Plan: set Sorxn = So

surr and solve for T;convert kJ to J

sysosys

o

surro ΔS

TΔH

ΔS

rxn implies system!!!

TΔSΔH

syso

syso

TJ/K 314

)J000,127(

Ans: T = +404 K

Gibbs Free Energy, G = H -TSGibbs Free Energy, G = H -TS

Multiply through by -TMultiply through by -T

-T∆S-T∆Sunivuniv = ∆H = ∆Hsyssys - T∆S - T∆Ssyssys

--T∆ST∆Sunivuniv = change in Gibbs free energy for the = change in Gibbs free energy for the

system = ∆Gsystem = ∆Gsystemsystem

hence, hence, ∆∆G = G = ∆∆H -TH -T∆∆SSUnder Under standard conditionsstandard conditions — —

∆∆GGoosyssys = ∆H = ∆Hoo

syssys - T∆S - T∆Soosyssys

syssys

univ Δ+ T

ΔH = ΔS S

∆∆SSunivuniv = ∆S = ∆Ssurrsurr + ∆S + ∆Ssyssys

J. Willard GibbsJ. Willard Gibbs1839-19031839-1903

∆∆G = ∆H - T∆SG = ∆H - T∆SGibbsGibbs free energyfree energy change =change = total energy total energy

change for system change for system

- energy lost in disordering the system- energy lost in disordering the system

If the reaction isIf the reaction is

•• exothermic (negative ∆H)

•• and entropy increases (positive ∆So)

•• thenthen ∆∆G G must bemust be NEGATIVE NEGATIVE

• the reaction is spontaneous (and product-favored) at ALL temperatures.

40

• G will be positive (∆G > 0) ∆G > 0) when H is positive (endothermic) and S is negative (more ordered). So the change in free energy will be positive at all temperatures.

• The reaction will therefore be nonspontaneous at ALL temperatures

• When G = 0 the reaction is at equilibrium

∆∆G = ∆H - T∆SG = ∆H - T∆S

41

G = H – TSSpontaneous or Not?

A decrease in Gibbs free energy (G < 0) corresponds to a spontaneous process

An increase in Gibbs free energy (G > 0) corresponds to a nonspontaneous process

Calculating ∆GCalculating ∆Goo : ∆G∆Goo = ∆∆H -T∆S∆Soo

Combustion of acetylene @ 25 oCC2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(g)

Use enthalpies of formation to calculate ∆Ho

rxn = -1238 kJUse standard molar entropies to calculate ∆So

rxn = -97.4 J/K or -0.0974 kJ/K

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

= -1209 kJ (spontaneous)Reaction is product-favored in spite of negative

∆Sorxn.

Reaction is “enthalpy driven”

From tables of thermodynamic data we findFrom tables of thermodynamic data we find

∆∆HHoorxnrxn = +25.7 kJ = +25.7 kJ (endothermic)(endothermic)

∆∆SSoorxnrxn = +108.7 J/K or +0.1087 kJ/K = +108.7 J/K or +0.1087 kJ/K (disorder)(disorder)

∆∆GGoorxnrxn = +25.7 kJ - (298 K)(+0.1087 kJ/K) = +25.7 kJ - (298 K)(+0.1087 kJ/K)

= -6.7 kJ = -6.7 kJ (spontaneous)(spontaneous)

Reaction is Reaction is product-favoredproduct-favored in spite of positive ∆H in spite of positive ∆Hoorxnrxn. .

Reaction is Reaction is “entropy driven”“entropy driven”

NHNH44NONO33(s) + (s) + heatheat NH NH44NONO33(aq)(aq)

Is the dissolution of ammonium nitrate product-Is the dissolution of ammonium nitrate product-favored? favored?

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

Calculating ∆GCalculating ∆Goo : ∆G∆Goo = ∆∆H -T∆S∆Soo

The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has H = +95.7 kJ and S = +142.2 J/K at 25°C.

Calculate G and determine if it is spontaneous.

Since G is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.

H = +95.7 kJ, S = 142.2 J/K, T = 298 K

G, kJ

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

GT, H, S

STHG

J 1033.5

2.142K 298J 1095.7

STHG

4

K

J3

The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has H = +95.7 kJ and S = +142.2 J/K.

Calculate the minimum T @ which it will be spontaneous.

The temperature must be higher than 673K for the reaction to be spontaneous (i.e. 674 K)

H = +95.7 kJ, S = 142.2 J/K, G < 0

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

TG, H, S

STHG

KJ3

KJ3

2.142TJ 1095.7

02.142TJ 1095.7

0 STHG

TK 673

T 2.142

J 1095.7

K

J

3

Gibbs Free Energy, GGibbs Free Energy, G

Calculating ∆GCalculating ∆Goo (two ways)

a)a) Determine ∆HDetermine ∆Hoorxnrxn and ∆S and ∆Soo

rxnrxn and and

use use Gibbs equation (at various temps).Gibbs equation (at various temps).

b)b) Use tabulated values of free

energies of formation, ∆Gfo

@ 25oC

∆∆GGoorxnrxn = = ∆G ∆Gff

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

rxnrxn = = ∆G ∆Gffoo (products) - (products) - ∆G ∆Gff

oo (reactants) (reactants)

47

Substance G°f kJ/mol

Substance G°f kJ/mol

Al(s) 0 Al2O3 -1576.5 Br2(l) 0 Br2(g) +3.14

C(diamond) +2.84 C(graphite) 0 CO(g) -137.2 CO2(g) -394.4 Ca(s) 0 CaO(s) -604.17 Cu(s) 0 CuO(s) -128.3 Fe(s) 0 Fe2O3(s) -740.98 H2(g) 0 H2O2(l) -120.4

H2O(g) -228.57 H2O(l) -237.13 HF(g) -270.70 HCl(g) -95.27

HBr(g) -53.22 HI(g) +1.30 I2(s) 0 I2(g) +19.37 N2(g) 0 NH3(g) -16.66 NO(g) +86.71 NO2(g) +51.84 Na(s) 0 O2(g) 0 S(s) 0 SO2(g) -300.4

Calculate G at 25C for the reactionCH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4O3(g)

standard free energies of formation from Appendix IIB

G, kJ

Solution:

Concept Plan:

Relationships:

Given:Find:

GGf of prod & react

reactantsproducts GGG frfp nn

kJ 3.148

)]kJ 0.0(8)kJ 5.50[(]kJ) 2.163(4)kJ 6.228(2)kJ 4.394[(

)]G(8)G[()]G()G(2)G[(

GGG

24322OCHOOHCO

reactantsproducts

fffff

frfp nn

Substance Gf, kJ/molCH4(g) -50.5O2(g) 0.0

CO2(g) -394.4H2O(g) -228.6O3(g) 163.2

(spontaneous)

The reaction SO2(g) + ½ O2(g) SO3(g) has H = -98.9 kJ and S = -94.0 J/K at 25°C.

Calculate G at 125C and determine if it is spontaneous.

Since G is -ve, the rxn is spontaneous at this temperature, but less spontaneous than at 25C (-127 kJ)

H = -98.9 kJ, S = -94.0 J/K, T = 398 K

G, kJ

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

GT, H, S STHG

kJ 5.61J 105.61

0.94K 398J 1098.9

STHG

3

K

J3

(PRACTICE PROBLEM)

∆∆G, ∆G˚, and KG, ∆G˚, and Keqeq

• ∆G is the change in free energy at non-standard conditions.

• ∆G is related to ∆G˚

• ∆G = ∆G˚ + RT ln Q where Q = reaction quotient

• When Q < K or Q > K, the reaction may be spontaneous or nonspontaneous.

• When Q = K reaction is at equilibrium

• When ∆G = 0 reaction is at equilibrium

• Therefore, ∆G˚ = - RT ln K

FACT: ∆Gorxn is the change in free energy when

pure reactants convert COMPLETELY to pure

products, both at standard conditions. FACT: Product-favored systems have

Keq > 1 (∆G˚∆G˚rxnrxn < 0) < 0).

Therefore, both ∆G˚Therefore, both ∆G˚rxnrxn and K and Keqeq are related are related

to reaction favorability.to reaction favorability.

Thermodynamics and KThermodynamics and Keqeq

Summary: Summary: ∆G˚ = - RT ln K

Calculate K for the reaction @ 25 Calculate K for the reaction @ 25 ooCC

NN22OO44 2 NO 2 NO22 ∆G∆Goorxnrxn = +4.8 kJ = +4.8 kJ

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

∆∆GGoorxnrxn = - RT lnK = - RT lnK

lnK = -4800 J

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

Thermodynamics and Thermodynamics and KKeqeq

K = eK = e–1.94 –1.94 = 0.14 = 0.14 (reactant favored)(reactant favored)

When ∆GWhen ∆Goorxnrxn > 0 > 0 (nonspontaneous),(nonspontaneous),

then K < 1!!then K < 1!!

• Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C

N2(g) + 3 H2(g) ⇄ 2 NH3(g)

G° = -RT lnK+46400 J = -(8.314 J/K)(700 K) lnK

ln K = -7.97K = e−7.97 = 3.45 10−4 small!!!

since K is << 1, the position of equilibrium favors reactants

H° = [ 2(-46.19)] − [ 0 +3( 0)] = -92.38 kJ = -92380 JS° = [2 (192.5)] − [(191.50) + 3(130.58)] = -198.2 J/K

G° = -92380 J - (700 K)(-198.2 J/K)

G° = +46400 J (nonspontaneous)

54

• Calculate G at 427°C for the reaction below if the PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0 atm

N2(g) + 3 H2(g) ⇄ 2 NH3(g)

Q = PNH3

2

PN21 PH2

3

(2.0 atm)2

(33.0 atm)1 (99.0)3= = 1.2 10-7

G = G° + RTlnQ spontaneous G = +46400 J + (8.314 J/K)(700 K)(ln 1.2 10-7)

H° = [ 2(-46.19)] - [0 +3( 0)] = -92.38 kJ = -92380 J

S° = [2 (192.5)] - [(191.50) + 3(130.58)] = -198.2 J/K

G° = -92380 J - (700 K)(-198.2 J/K)

G° = +46400 J (nonspontaneous)

G = 46400 J − 92700 J = -46300 J = −46 kJ −G°

55

Q. Rank the following in order of increasing molar entropy (So) @ 25oC!

a) Cl2(g), I2(g), Br2(g), and F2(g)

b) H2O(g), H2O2 (g), H2S(g)

F2(g) < Cl2(g) < Br2(g) < I2(g)

H2O(g) < H2S(g) < H2O2(g)

56

2 22NO(g) O (g) 2NO (g)

C) Ssurr = +114 kJ/K, reaction is spontaneous

A) Ssurr = +114 kJ/K, reaction is not spontaneous

B) Ssurr = +321 J/K, reaction is spontaneous

D) Ssurr = -355 J/K, reaction is not spontaneous

E) Ssurr = +321 J/K, it is not possible to predict the spontaneity of this reaction without more information.

Use only the information provided here to determine the value of Ssurr @ 355 K. Predict whether this reaction shown will be spontaneous @ this temperature, if Hrxn = -114 kJ.

G under Nonstandard ConditionsG = G only when the reactants and products

are in their standard statesthere normal state at that temperaturepartial pressure of gas = 1 atmconcentration = 1 M

under nonstandard conditions, G = G + RTlnQQ is the reaction quotient

at equilibrium G = 0G = − RTlnK and G° = H − T S°

H − TS° = − RTlnK, by rearranging

RTlnK = −H − TS°, and dividing by R T

Temperature Dependence of K

• for an exothermic reaction, increasing the temperature decreases the value of the equilibrium constant

• for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant

R

S

T

1

R

Hln rxnrxn

K

RTlnK −H TS°———— = ———— + ———— RT RT R T