Chapter 17Free Energy and Thermodynamics

Chemistry II

First Law of Thermodynamics

• First Law of Thermodynamics: Energy cannot be Created or Destroyed the total energy of the universe cannot change though you can transfer it from one place to another

ΔEuniverse = 0 = ΔEsystem + ΔEsurroundings

First Law of Thermodynamics

• Conservation of Energy• For an exothermic reaction, “lost” heat from the system goes into the

surroundings

• two ways energy “lost” from a system, converted to heat, q used to do work, w

• Energy conservation requires that the energy change in the system equal the heat released + work done

Δ E = q + w

Energy Tax

• you can’t break even!• to recharge a battery with 100 kJ of useful

energy will require more than 100 kJ • every energy transition results in a “loss” of

energy conversion of energy to heat which is “lost” by

heating up the surroundings

e.g. if a light bulb is 5 % efficient 95% is lost as heat

Heat Tax

fewer steps generally results in a lower total heat

tax

Spontaneous and Nonspontaneous Processes

• e.g. in physics - will an object fall when dropped? in chemistry - will iron rust?

• thermodynamics predicts whether a process will proceed under the given conditions - spontaneous process

nonspontaneous processes require energy inputs

• In physics sponaneity is determined by looking at the potential energy change of a system

• In chemistry spontaneity is determined by comparing the free energy change of the system if the system after reaction has less free energy than before the

reaction, the reaction is thermodynamically favorable.

• spontaneity ≠ fast or slow

Comparing Potential Energy

The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.

High PE → Low PE

High ‘Free Energy’ → Low ‘Free Energy’

Reversibility of Process

• any spontaneous process is irreversible it will proceed in only one direction

• a reversible process will proceed back and forth between the two end conditions equilibrium results in no change in free energy

• if a process is spontaneous in one direction, it must be nonspontaneous in the opposite direction

Thermodynamics vs. Kinetics

Don’t confuse spontaneity (direction and extent of reaction) with speed

This reaction will require an energy input since ‘free energy’ of reactants < ‘free energy’ products

Diamond → Graphite

Graphite has less FE than diamond, so the conversion of diamond into graphite is spontaneous – but don’t worry, it’s so slow that your ring won’t turn into pencil lead in your lifetime (or

through many of your generations).

Factors Affecting Whether a Reaction Is Spontaneous

• The two factors that determine spontaneity (free energy) 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. Δ H

• The entropy factors relates to the randomness/orderliness of a system ΔS

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

Enthalpy, ΔH• ΔH related to the internal energy (kJ/mol)

• stronger bonds = more stable molecules

• if products more stable than reactants, energy released Exothermic (ΔH = negative)

• if reactants more stable than products, energy absorbed Endothermic (ΔH = positive)

Hess’ Law: Δ H°rxn = Σ(nΔ H°prod) - Σ(nΔ H°react)

Where n = stoichiometric coefficient

Changes in Entropy, ΔS• entropy change is favorable when the result is a more random system

ΔS is positive

• Some changes that increase the entropy are: reactions whose products are in a more disordered state.

(solid > liquid > gas) reactions which have larger numbers of product molecules than

reactant molecules increase in temperature solids dissociating into ions upon dissolving

Increases in EntropyMelting

Evaporating

Dissolution

Entropy• entropy is a thermodynamic function that increases as the number of

energetically equivalent ways of arranging the components increases, a measure of randomness, S S generally J/K

• S = k ln W k = Boltzmann Constant = 1.38 x 10-23 J/K W is the number of energetically equivalent ways, unitless

• A chemical system proceeds in a direction that increases the entropy of the universe

• Next we consider W,

W

Energetically Equivalent States for the Expansion of a Gas in a vacuum

The following are macro states

Macrostates → Microstates

This macrostate can be achieved throughseveral different arrangements of the particles

These microstates all have the same

macrostate

So there are 6 different particle

arrangements that result in the same

macrostate

Macrostates and ProbabilityThere is only one possible

arrangement that gives State A and one that gives State B

There are 6 possible arrangements that give State C

Therefore State C has higher entropy than either State A or

State B

The macrostate with the highest entropy also has the greatest

dispersal of energy, now imagine thousands of atoms!

Entropy• Second law of thermodynamics:

For any spontaneous process, the entropy of the universe increases, ΔSuniv > 0

• Spontaneous processes increase the entropy of the universe

Δ S = Sfinal – Sinitial

• In our flask analogy, ΔS > 0

• Explains why heat travels from a substance at high temp. to a substance at low temp. but never the opposite

Entropy Change in State Change

• when materials change state, the number of macrostates it can have changes as well for entropy: solid < liquid < gasbecause the degrees of freedom of motion increases

solid → liquid → gas

Entropy Change and State Change

Heat Flow, Entropy, and the 2nd Law

Heat must flow from water to ice in order for the entropy of the universe to increase

Is Entropy Increases or Decreasing?

• Iodine sublimes

• A liquid boils

• A solid decomposes into two gaseous products

Heat Transfer and Changes in S of Surroundings

• the total entropy change of the universe must be positive for a process to be spontaneous for reversible process ΔSuniv = 0, for irreversible (spontaneous) process Δ Suniv > 0

ΔSuniverse = ΔSsystem + Δ Ssurroundings

• if the entropy of the system decreases, then the entropy of the surroundings must increase by a larger amount when ΔSsystem is negative, ΔSsurroundings is positive e.g. when water freezes

• the increase in ΔSsurroundings often comes from the heat released in an exothermic reaction

Temperature Dependence of ΔSsurroundings

• when a system process is exothermic, it adds heat to the surroundings, increasing the entropy of the surroundings

• when a system process is endothermic, it takes heat from the surroundings, decreasing the entropy of the surroundings

• the amount the entropy of the surroundings changes depends on the temperature it is at originally the higher the original temperature, the less effect addition or

removal of heat has

TH

S systemgssurroundin

Temperature Dependence of ΔSsurroundings

• Example:2N2 (g) + O2 (g) → 2N2O (g) ΔH = +163.2 kJ/mol

(i) Calculate the entropy change in the surroundings at 25 ºC

ΔSsurr. = -ΔHsys. / T = -163.2 kJ/298K = -0.548 kJ/K = -548 J/K

(ii) Determine the sign of the entropy change

3 mols on LHS, 2 mols on RHS so -ΔSsys

(iii) Determine the sign of Δ Suniverse , is reaction spontaneous?

Δ Suniverse = ΔSsystem + Δ Ssurroundings

-ve + -ve = -veNot spontaneous

Gibbs Free Energy, ΔG

• Enthalpy (ΔHsys) and entropy (ΔSsurr) are related by:

ΔSsurr. = -ΔHsys. / T

• Also: Δ Suniverse = ΔSsystem + Δ Ssurroundings

• Combining these: ΔSuniverse = ΔSsystem -ΔHsys. / T

• Now ΔSuniv can be calculated with the ‘system’ only

• Multiplying by –T: -TΔSuniverse = -TΔSsystem -ΔHsys.

• This is now called the Gibbs free Energy expression (remember we multiplied by a –ve)

Gibbs Free Energy, ΔG

• maximum amount of energy from the system available to do work on the surroundings

G = H – T∙S

ΔGsys = Δ Hsys – T Δ Ssys

The change in Gibbs free energy for a process at constant T and P is ~ -ΔSuniv

Δ Gsys = – T Δ Suniverse

• ΔG (Gibbs free energy) being –ve is a criterion for spontaneity• there is a decrease in free energy of the system that is released into the

surroundings; therefore a process will be spontaneous when ΔG is negative (i.e. ΔS +ve, since we mult. by - T)

Gibbs Free Energy, ΔG

• process will be spontaneous when ΔG is negative

ΔGsys = Δ Hsys – T Δ Ssys

Three possibilities:

(1) Δ G will be negative - spontaneous Δ H is negative and ΔS is positive

exothermic and more random

ΔH is negative and large and ΔS is negative but small

Δ H is positive but small and ΔS is positive high temperature favors spontaneity, nonspon. at low T

(2) ΔG will be +ve when ΔH is +ve and ΔS is −ve - nonspontaneous never spontaneous at any temperature

(3) when ΔG = 0 the reaction is at equilibrium

Which of the following is spontaneous?

ΔHº TΔSº

ΔHº

TΔSº

ΔHº

TΔSº

ΔHº TΔSº

ΔG, ΔH, and ΔS

Ex. 17.3a – The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has ΔH = +95.7 kJ and Δ S = +142.2 J/K at 25°C.

Calculate Δ G and determine if it is spontaneous.

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

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

ΔG, kJ

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

Δ GT, ΔH, ΔS

STHG

J 1033.5

2.142K 298J 1095.7

STHG

4KJ3

Ex. 17.3a – The reaction CCl4(g) C(s, graphite) + 2 Cl2(g) has ΔH = +95.7 kJ and ΔS = +142.2 J/K.

Calculate the minimum temperature it will be spontaneous.

The temperature must be higher than 673K for the reaction to be spontaneous

ΔH = +95.7 kJ, ΔS = 142.2 J/K, ΔG < 0

T, K

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

TΔG, ΔH, ΔS

STHG

K

J3

KJ3

2.142TJ 1095.7

02.142TJ 1095.7

0 STHG

K 673T

T 2.142

J 1095.7

KJ

3

Entropy• entropy is a thermodynamic function that increases as the number of

energetically equivalent ways of arranging the components increases, a measure of randomness, S S generally J/K

• S = k ln W k = Boltzmann Constant = 1.38 x 10-23 J/K W is the number of energetically equivalent ways, unitless

• A chemical system proceeds in a direction that increases the entropy of the universe

• Next we consider W,

The 3rd Law of ThermodynamicsAbsolute Entropy

• the absolute entropy of a substance is the amount of energy it has due to dispersion of energy through its particles

• the 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K therefore, every substance that is not a

perfect crystal at absolute zero has some energy from entropy

therefore, the absolute entropy of substances is always +ve

Standard Entropies

• S° (J/mol.K)

• Extensive property (depends on amount)

• 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

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

Relative Standard EntropiesStates

• the gas state has a larger entropy than the liquid state at a particular temperature

• the liquid state has a larger entropy than the solid state at a particular temperature

Substance S°, (J/mol∙K)

H2O (l) 70.0

H2O (g) 188.8

Relative Standard EntropiesMolar Mass

• the larger the molar mass, the larger the entropy

Relative Standard EntropiesAllotropes

• the less constrained the structure of an allotrope is, the larger its entropy

Relative Standard EntropiesMolecular Complexity

• larger, more complex molecules have larger entropy than single atoms

• more available energy states (translational, rotational, vibrational) allowing more dispersal of energy through the states

SubstanceMolarMass

S°, (J/mol∙K)

Ar (g) 39.948 154.8

NO (g) 30.006 210.8

Relative 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

Standard Entropy of Reaction, ΔSrxn

• Calculated similar to ΔHrxn

• Where n = stoichiometric coefficient

reactantsproducts SSS rp nn

Ex. 17.4 –Calculate ΔS for the reaction4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(l)

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

standard entropies from Appendix IIBΔS, J/K

Check:

Solution:

Concept Plan:

Relationships:

Given:Find:

ΔSSNH3, SO2, SNO, SH2O,

reactantsproducts SSS rp nn

KJ

KJ

KJ

KJ

KJ

)(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/molKNH3(g) 192.8

O2(g) 205.2

NO(g) 210.8H2O(g) 188.8

Calculating ΔGrxn

• At 25 C and at temperatures other than 25 C: assuming the change in ΔHo

reaction and Δ Soreaction are small over a

limted temp. range

Δ Grxn = Δ Hrxn – T Δ Srxn

Ex. 17.6 – The reaction SO2(g) + ½ O2(g) SO3(g) has ΔHrxn = -98.9 kJ and ΔSrxn = -94.0 J/K at 25°C.

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

Since ΔG is -, the reaction is spontaneous at this temperature, though less so than at 25C

ΔH = -98.9 kJ, ΔS = -94.0 J/K, T = 398 K

ΔG, kJ

Answer:

Solution:

Concept Plan:

Relationships:

Given:

Find:

ΔGT, ΔH, ΔS STHG

kJ 5.61J 105.61

0.94K 398J 1098.9

STHG

3KJ3

Calculating ΔGrxn with ΔG f

• at 25C:ΔGo

rxn = ΣnpΔGof(products) - ΣnrΔGo

f(reactants)

• Like ΔHf°, ΔGf° = 0 for an element in its standard state:

• Example: ΔGf° (O2(g)) = 0

52

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

Ex. 17.7 –Calculate ΔG at 25C for the reactionCH4(g) + 8 O2(g) CO2(g) + 2 H2O(g) + 4 O3(g)

standard free energies of formation from Appendix IIBΔG, kJ

Solution:

Concept Plan:

Relationships:

Given:Find:

ΔGΔGf of prod & react

reactantsproducts GGG frfp nn

kJ 3.148)]kJ 0.0(8)kJ 5.50[(]kJ) 2.163()kJ 6.228(2)kJ 4.394[(

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

GGG

24322 OCHOOHCO

reactantsproducts

fffff

frfp nn

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

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

ΔG Relationships

• if a reaction can be expressed as a series of reactions, the sum of the ΔG values of the individual reaction is the ΔG of the total reaction ΔG is a state function

• if a reaction is reversed, the sign of its ΔG value reverses

• if the amounts of materials is multiplied by a factor, the value of the ΔG is multiplied by the same factor the value of ΔG of a reaction is extensive

ΔG°rxn ≠ Reaction Rate

• Although ΔG°rxn can be used to predict if a reaction will be spontaneous as written, it does not tell us how fast a reaction will proceed.

• Example:

C(s, diamond) + O2(g) CO2(g) ΔG°rxn= -397 kJ<<0

But diamonds are forever…. ΔG°rxn≠ rate

Why Free Energy is ‘Free’

• If ΔG –ve: change in Gibbs free energy = max amount of available ‘free’ energy to do work, should be called ‘Gibbs available energy’

• For many reactions ΔG < ΔH e.g.

C(s) + 2H2(g) →CH4(g)

ΔHrxn = -74.6 kJ, ΔSºrxn = -80.8 J/K, ΔGºrxn = -50.5 kJ

• Exothermic reaction but only half as much available to do work

• Entropy is –ve, so some of emitted heat must inc. entropy of the universe in order to make ΔGºrxn -ve

Free Energy and Reversible Reactions

• the change in free energy is a theoretical limit as to the amount of work that can be done

• if the reaction achieves its theoretical limit, it is theoretically a reversible reaction

Real Reactions

• in a real reaction, some of the free energy is “lost” as heat if not most

• therefore, real reactions are irreversible

Energy is lost – when the battery does work on the motor some energy is lost in the wires as heat (resistance to e- flow)

ΔG under Nonstandard Conditions Δ G = Δ G only when the reactants and products are in their standard

states there normal state at that temperaturepartial pressure of gas = 1 atmconcentration = 1 M

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

at equilibrium ΔG = 0ΔG = ─RTlnK

ΔG under Nonstandard Conditions Spilled water spontaneously evaporates even though ΔG◦rxn for the

vaporization of water is +ve..why?

We do not live at std. conditions

Q = PH2O

Under std. conditions water does not vaporize

ΔG rxn = ΔG◦rxn

= +ve

Spontaneous in reverse direction

Water condenses

ΔG◦rxn = +8.56 kJ/mol

ΔG rxn = -ve

Non-std.Conditions

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

atmN2(g) + 3 H2(g) → 2 NH3(g)

Q = PNH32

PN21 x PH2

3

(2.0 atm)2

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

ΔG = ΔG° + RTlnQ

ΔG = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)

ΔG = -46300 J = -46 kJ

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

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

ΔG° = -92380 J - (700 K)(-198.2 J/K)

ΔG° = +46400 J

Free Energy and Equilibrium:Relating ΔG°rxn to K

• Standard free energy change of a reaction and the equilibrium constant are related

• K becomes larger as ΔG° becomes more –ve

• If ΔG° is large and –ve then reaction has a large equil. constant, if ΔG° is large and +ve reaction has a small equil. Constant, favors reactants at equilibrium

• ΔG = ΔG + RTlnQ and at equilibrium Q = K, ΔG = 0

ΔG°rxn = -RT lnK

Free Energy and Equilibrium:Relating ΔG°rxn to K

• ΔGrxn = ΔGrxn + RT lnQ and at equilibrium Q = K, ΔGrxn = 0

ΔG°rxn = -RT lnK

• When K < 1, lnK is –ve and ΔG°rxn is +ve When Q = 1, reaction is spontaneous in reverse direction

• When K > 1, lnK is +ve and ΔG°rxn is -ve When Q = 1, reaction is spontaneous in forward direction

• When K = 1, lnK is 0 and ΔG°rxn is 0 Reaction is at equilibrium under std. conditions

Example - K

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

N2(g) + 3 H2(g) 2 NH⇌ 3(g)

ΔG° = -RT lnK

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

lnK = -7.97

K = e-7.97 = 3.45 x 10-4

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

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

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

ΔG° = -92380 J - (700 K)(-198.2 J/K)

ΔG° = +46400 J