Chapter 19 thermodynamics
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Chapter 19: Thermodynamics. Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop. Thermodynamics. Study of energy changes and flow of energy Answers several fundamental questions: Is it possible for a given reaction to occur?

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Chapter 19: Thermodynamics

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Chapter 19 thermodynamics

Chapter 19: Thermodynamics

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop


Thermodynamics

Thermodynamics

  • Study of energy changes and flow of energy

  • Answers several fundamental questions:

    • Is it possible for a given reaction to occur?

    • Will the reaction occur spontaneously (without outside interference) at a given T?

    • Will reaction release or absorb heat?

  • Tells us nothing about time frame of reaction

    • Kinetics

  • Two major considerations

  • Enthalpy changes, H (heats of reaction)

    • Heat exchange between system and surroundings

  • Nature's trend to randomness or disorder

    • Entropy


Review of first law of thermodynamics

Review of First Law of Thermodynamics

  • Internal energy, E

    • System's total energy

    • Sum of KE and PE of all particles in system

    • or for chemical reaction

    • E +energy into system

    • E–energy out of system


Two methods of energy exchange between system and surroundings

Two Methods of Energy Exchange Between System and Surroundings

  • HeatqWorkw

  • E = q + w

  • Conventions of heat and work


First law of thermodynamics

First Law of Thermodynamics

  • Energy can neither be created nor destroyed

  • It can only be converted from one form to another

    • Kinetic  Potential

    • Chemical  Electrical

    • Electrical  Mechanical

  • E is a state function

    • E is a change in a state function

    • Path independent

    • E = q + w


Work in chemical systems

Work in Chemical Systems

  • Electrical

  • Pressure-volume orPV

    • w = –PV

      • Where P = external pressure

    • If PV only work in chemical system, then


Heat at constant volume

Heat at Constant Volume

  • Reaction done at constantV

    • V = 0

    • PV = 0, so

    • E = qV

  • Entire energy change due to heat absorbed or lost

  • Rarely done, not too useful


Heat at constant pressure

Heat at Constant Pressure

  • More common

  • Reactions open to atmosphere

    • Constant P

  • Enthalpy

    • H = E + PV

  • Enthalpy change

    • H = E + PV

  • Substituting in first law for E gives

    • H = (q –PV)+ PV = qP

    • H = qP

      • Heat of reaction at constant pressure


Converting between e and h for chemical reactions

Converting Between E and H For Chemical Reactions

  • H E

  • Differ byH– E = PV

  • Only differ significantly when gases formed or consumed

  • Assume gases are ideal

  • Since P and T are constant


Converting between e and h for chemical reactions1

Converting Between E and H For Chemical Reactions

  • When reaction occurs

    • V caused by n of gas

  • Not all reactants and products are gases

    • So redefine as ngas

    • Where ngas= (ngas)products– (ngas)reactants

  • Substituting into H = E + PV gives

    • or


Chapter 19 thermodynamics

Ex. 1 What is the difference between H and E for the following reaction at 25 °C?

2 N2O5(g) 4 NO2(g) + O2(g)

What is the % difference between H and E?

Step 1: Calculate H using data (Table 7.2)

Recall

H° = (4 mol)(33.8 kJ/mol) +

(1 mol)(0.0 kJ/mol) – (2 mol)(11 kJ/mol)

H° = 113 kJ


Ex 1 h and e cont

Ex. 1. H and E (cont.)

Step 2: Calculate ngas

ngas = (ngas)products–(ngas)reactants

ngas = (4 + 1 – 2) mol = 3 mol

Step 3: Calculate E using

R = 8.31451 J/K molT = 298 K

E = 113 kJ – (3 mol)(8.314 J/K mol)(298 K)(1 kJ/1000 J)

E = 113 kJ – 7.43 kJ = 106 kJ


Ex 1 h and e cont1

Ex. 1. H and E (cont.)

Step 4: Calculate percent difference

Bigger than most, but still small

Note: Assumes that volumes of solids and liquids are negligible

VsolidVliquid << Vgas


Is assumption that v solid v liquid v gas justified

Is Assumption that VsolidVliquid << Vgas Justified?

  • Consider

    CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O+ CO2(g)

    37.0 mL 2×18.0 mL 18 mL 18 mL 24.4 L

  • Volumes assuming each coefficient equal number of moles

  • So V = Vprod– Vreac = 24.363 L  24.4 L

  • Yes, assumption is justified

    Note: If no gases are present reduces to


Learning check

Learning Check

  • Consider the following reaction for picric acid:

    8O2(g) + 2C6H2(NO2)3OH(l)→ 3 N2(g) + 12CO2(g) + 6H2O(l)

  • Calculate Η°, Ε°

H° = 12 mol(–393.5 kJ/mol) + 6 mol(–241.83 kJ/mol) + 6 mol(0.00 kJ/mol) – 8 mol(0.00 kJ/mol) – 2 mol(3862.94 kJ/mol)

ΔH° = –13,898.9 kJ

Ε° = H° – ngasRT = H° – (15 – 8) mol × 298 K × 8.314 × 10–3 kJ/(mol K)

Ε° = –13,898.9 kJ – 29.0 kJ =–13,927.9 kJ


Your turn

Your Turn!

Given the following:

3H2(g) + N2(g) → 2NH3(g) Η°= –46.19 kJ mol–1

Determine E for the reaction.

A. –51.14 kJ mol–1

B. –41.23 kJ mol–1

C. –46.19 kJ mol–1

D. –46.60 kJ mol–1

  • Η = E + nRTE = Η – nRT

  • E = –46.19 kJ mol –

    (–2 mol)(8.314 J K–1mol–1)(298 K)(1 kJ/1000 J)

  • E = –51.14 kJ mol–1


Enthalpy changes and spontaneity

Enthalpy Changes and Spontaneity

  • What are relationships among factors that influence spontaneity?

  • Spontaneous Change

    • Occurs by itself

    • Without outside assistance until finished

  • e.g.

    • Water flowing over waterfall

    • Melting of ice cubes in glass on warm day


Nonspontaneous change

Nonspontaneous Change

  • Occurs only with outside assistance

  • Never occurs by itself:

    • Room gets straightened up

    • Pile of bricks turns into a brick wall

    • Decomposition of H2O by electrolysis

  • Continues only as long as outside assistance occurs:

    • Person does work to clean up room

    • Bricklayer layers mortar and bricks

    • Electric current passed through H2O


Nonspontaneous change1

Nonspontaneous Change

  • Occur only when accompanied by some spontaneous change

    • You consume food, spontaneous biochemical reactions occur to supply muscle power

      • to tidy up room or

      • to build wall

    • Spontaneous mechanical or chemical change to generate electricity


Direction of spontaneous change

Direction of Spontaneous Change

  • Many reactions which occur spontaneously are exothermic:

    • Iron rusting

    • Fuel burning

  • H and E are negative

    • Heat given off

    • Energy leaving system

  • Thus, H is one factor that influences spontaneity


Direction of spontaneous change1

Direction of Spontaneous Change

  • Some endothermicreactions occur spontaneously:

    • Ice melting

    • Evaporation of water

    • Expansion of CO2 gas into vacuum

  • H and E are positive

    • Heat absorbed

    • Energy entering system

  • Clearly other factors influence spontaneity


Your turn1

Your Turn!

We can expect the combustion of propane

to be:

A. spontaneous

B. non-spontaneous

C. neither


Entropy symbol s

Entropy (Symbol S)

  • Thermodynamic quantity

  • Describes number of equivalent ways that energy can be distributed

  • Quantity that describes randomness of system

  • Greater statistical probability of particular state means greater the entropy!

    • Larger S, means more possible ways to distribute energy and that it is a more probable result


Fig 19 6 entropy distribution

Fig. 19.6 - Entropy Distribution

  • Low Entropy – (a)

    • A absorbs E in units of 10

    • Few ways to distribute E

    • ●represent E’s of molecules of A

  • High Entropy – (b)

    • More ways to distribute E

    • B absorbs E in units of 5

    • ●represent E’s of molecules of B


Entropy

Entropy

  • If Energy = money

  • Entropy (S) describes number of different ways of counting it


Examples of spontaneity

Examples of Spontaneity

  • Spontaneous reactions

    • Things get rusty spontaneously

      • Don't get shiny again

    • Sugar dissolves in coffee

      • Stir more—it doesn't undissolve

    • Ice  liquid water at RT

      • Opposite does NOT occur

    • Fire burns wood, smoke goes up chimney

      • Can't regenerate wood

  • Common factor in all of these:

    • Increase in randomness and disorder of system

    • Something that brings about randomness more likely to occur than something that brings order


Entropy s

Entropy, S

  • State function

  • Independent of path

  • S = Change in entropy

  • For chemical reactions or physical processes


Effect of volume on entropy

Effect of Volume on Entropy

  • For gases, entropy increases as volume increases

    • Gas separated from vacuum by partition

    • Partition removed, more ways to distribute energy

    • Gas expands to achieve more probable particle distribution

      • More random, higher probability, more positive S


Effect of temperature on entropy

Effect of Temperature on Entropy

  • As Tincreases, entropy increases

    (a) T = 0 K, particles in equilibrium lattice positions and S relatively low

    (b) T> 0 K, molecules vibrate, Sincreases

    (c) Tincreases further, more violent vibrations occur and Shigher than in (b)


Effect of physical state on entropy

Effect of Physical State on Entropy

  • Crystalline solid very low entropy

  • Liquid higher entropy,molecules can move freely

    • More ways to distribute KE among them

  • Gas highest entropy,particles randomly distributed throughout container

    • Many, many ways to distribute KE


Entropy affected by number of particles

Entropy Affected by Number of Particles

  • Adding particles to system

  • Increase number of ways energy can be distributed in system

  • So all other things being equal

  • Reaction that produces more particles will have positive S


Your turn2

Your Turn!

Which represents an increase in entropy?

A. Water vapor condensing to liquid

B. Carbon dioxide subliming

C. Liquefying helium gas

D. Proteins forming from amino acids


Entropy changes in chemical reactions

Entropy Changes in Chemical Reactions

Reactions without gases

  • Calculate number of mole molecules

    n = nproducts – nreactants

    • If n is positive, entropy increases

    • More molecules, means more disorder

    • Usually the side with more molecules, has less complex molecules

Reactions involving gases

  • Calculate change in number of moles of gas, ngas

  • If ngas is positive , S is positive

  • ngas is more important than nmolecules


Entropy changes in chemical reactions1

Entropy Changes in Chemical Reactions

Ex. N2(g) + 3H2(g) 2NH3(g)

nreactant = 4nproduct = 2

n = 2 – 4 = –2

Predict Srxn < 0

Lower positional probability

Higher positional probability


Ex 2 predict sign of s for following reactions

Ex. 2 Predict Sign of S for Following Reactions

CaCO3(s) + 2H+(aq) Ca2+(aq) + H2O + CO2(g)

  • ngas = 1 mol – 0 mol = 1 mol

  • sincengas is positive, S is positive

    2 N2O5(g) 4 NO2(g) + O2(g)

  • ngas = 4 mol + 1 mol – 2 mol = 3 mol

  • sincengas is positive, S is positive

    OH–(aq) + H+(aq) H2O

  • ngas = 0 mol

  • n = 1 mol – 2 mol = –1 mol

  • sincengas is negative, S is negative


Predict sign of s in the following

Predict Sign of S in the Following:

Dry ice → carbon dioxide gas

Moisture condenses on a cool window

AB → A + B

A drop of food coloring added to a glass of water disperses

2Al(s) + 3Br2(l) → 2AlBr3(s)

CO2(s) → CO2(g)

positive

H2O(g) → H2O(l)

negative

positive

positive

negative


Your turn3

Your Turn!

Which of the following has the most entropy at standard conditions?

  • H2O(l)

  • NaCl(aq)

  • AlCl3(s)

  • Can’t tell from the information


Your turn4

Your Turn!

Which reaction would have a negative entropy?

A.Ag+(aq) + Cl–(aq) → AgCl(s)

B.N2O4(g) → 2NO2(g)

C. C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(g)

D. CaCO3(s) → CaO(s) + CO2(g)


Both entropy and enthalpy affect reaction spontaneity

Both Entropy and Enthalpy Affect Reaction Spontaneity

  • Sometimes they work together

    • Building collapses

    • PE decreases His negative

    • Stones disordered S is positive

  • Sometimes work against each other

    • Ice melting (ice/water mix)

    • Endothermic

      • H is positive nonspontaneous

    • Increase in disorder of molecules

      • S is positive spontaneous


Which prevails

Which Prevails?

  • Hard to tell—depends on temperature!

    • At 25 °C, ice melts

    • At –25 °C, water freezes

  • So three factors affect spontaneity:

    • H

    • S

    • T

    • Next few slides will develop the relationship between H, S,and T that defines a spontaneous process


Second law of thermodynamics

Second Law of Thermodynamics

  • When a spontaneous event occurs, total entropy of universe increases

    • (Stotal > 0)

  • In a spontaneous process, Ssystem can decrease as long as total entropy of universe increases

    • Stotal = Ssystem + Ssurroundings

  • It can be shown that


Law of conservation of energy

Spontaneous Reactions (cont.)

Law of Conservation of Energy

  • Says q lost by system must be gained by surroundings

    • qsurroundings= –qsystem

  • If system at constant P, then

    • qsystem = H

  • So

    • qsurroundings = –Hsystem

  • and


Thus entropy for entire universe is

Thus Entropy for Entire Universe is

Multiplying both sides by T we get

TStotal = TSsystem – Hsystem

or

TStotal = – (Hsystem– TSsystem)

  • For reaction to be spontaneous

    • TStotal > 0 (entropy must increase)

      So,

      (Hsystem– TSsystem) < 0

      must be negative for reaction to be spontaneous


Gibbs free energy

Gibbs Free Energy

  • Would like one quantity that includes all three factors that affect spontaneity of a reaction

  • Define new state function, G

  • Gibbs Free Energy

    • Maximum energy in reaction that is "free" or available to do useful work

      GH – TS

  • At constant P and T, changes in free energy G = H – TS


Gibbs free energy1

G = H – TS

G  state function

Made up of T, H and S = state functions

Has units of energy

Extensive property

G = Gfinal– Ginitial

Gibbs Free Energy


Criteria for spontaneity

Criteria for Spontaneity?

  • At constant P and T, process spontaneous only if it is accompanied by decrease in free energy of system


Summary

Summary

  • When H and S have same sign, T determines whether spontaneous or nonspontaneous

  • Temperature-controlledreactions are spontaneous at one temperature and not at another


Your turn5

Your Turn!

At what temperature (K) will a reaction become nonspontaneous when H = –50.2 kJ mol–1 and S = +20.5 J K–1 mol–1?

A. 298 K

B. 1200 K

C. 2448 K

D. The reaction cannot become non-spontaneous at any temperature


Third law of thermodynamics

Third Law of Thermodynamics

  • At absolute zero (0 K)

    • Entropy of perfectly ordered, pure crystalline substance is zero

  • S = 0 at T = 0 K

  • Since S = 0 at T = 0 K

    • Define absolute entropy of substance at higher temperatures

  • Standard entropy, S°

    • Entropy of 1 mole of substance at 298 K (25 °C) and 1 atm pressure

    • S° = S for warming substance from 0 K to 298 K (25 °C)


Consequences of third law

Consequences of Third Law

  • All substances have positive entropies as they are more disordered than at 0 K

    • Heating increases randomness

    • S° is biggest for gases—most disordered

  • For elements in their standard states

    • S°  0 (but Hf° = 0)

  • Units of S°  J/(mol K)

    Standard Entropy Change

    • To calculate S° for reaction, do Hess's Law type calculation

    • Use S° rather than entropies of formation


Learning check1

Calculate S° for the following:

CO2(s)→CO2(g)

S°: 187.6213.7 J/molK

S° = (213.7 – 187.6) J/molK

S° = 26.1 J/molK

CaCO3(s)→ CO2(g) + CaO(s)

S°: 92.9  213.7 40 J/molK

S° = (213.7 +40 – 92.9) J/molK

S° = 161 J/molK

Learning Check


Chapter 19 thermodynamics

Ex. 3.Calculate S° for reduction of aluminum oxide by hydrogen gas

Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)


Ex 3 calculate s cont

Ex. 3 Calculate ΔS° (cont.)

S° = 56.5 J/K + 566.1 J/K – (51.00 J/K + 391.8 )

S° = 179.9 J/K


Your turn6

Your Turn!

What is the entropy change for the following

reaction?

Ag+(aq) + Cl–(aq) → AgCl(s)

So 72.68 56.5 96.2 J K–1 mol–1

A. +32.88 J K–1 mol–1

B. –32.88 J K–1 mol–1

C. –32.88 J mol–1

D. +112.38 J K–1 mol–1

S° = [96.2 – (72.68 + 56.5)] J K–1 mol–1

S° = –32.88 J K–1 mol–1


Standard free energy changes

Standard Free Energy Changes

  • Standard Free Energy Change, G°

    • G measured at 25 °C (298 K) and 1 atm

  • Two ways to calculate, depending on what data is available

  • Method 1:

  • G° = H°– TS°

  • Method 2:


Ex 4 calculate g method 1

Ex. 4. Calculate G° Method 1

Calculate G ° for reduction of aluminum oxide by hydrogen gas

Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

Step 1: Calculate H° for reaction using heats of formation below


Ex 4 calculate g method 11

Ex. 4. Calculate G° Method 1

H ° = 0.0 kJ – 725.4 kJ – 0.00 kJ – (– 1669.8 kJ)

H ° = 944.4 kJ


Ex 4 calculate g method 12

Ex. 4. Calculate G° Method 1

Step 2:Calculate S°  see Example 3

S° = 179.9 J/K

Step 3:Calculate G° = H°– (298.15 K)S°

G° = 944.4 kJ – (298 K)(179.9 J/K)(1 kJ/1000 J)

G° = 944.4 kJ – 53.6 kJ = 890.8 kJ

  • G° is positive

  • Indicates that the reaction isnot spontaneous


Ex 4 calculate g method 2

Ex. 4. Calculate G° Method 2

  • Use Standard Free Energies of Formation

  • Energy to form 1 mole of substance from its elements in their standard states at 1 atm and 25 °C


Ex 4 calculate g method 21

Ex. 4. Calculate G° Method 2

Calculate G° for reduction of aluminum oxide by hydrogen gas.

Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)


Ex 4 calculate g method 22

Ex. 4. Calculate G° Method 2

G° = 0.0 kJ – 685.8 kJ – 0.00 kJ

– (– 1576.4 kJ)

G°= 890.6 kJ

Both methods same within experimental error


Spontaneous reactions produce useful work

Spontaneous Reactions Produce Useful Work

  • Fuels burned in engines to power cars or heavy machinery

  • Chemical reactions in batteries

    • Start cars

    • Run cellular phones, laptop computers, mp3 players

  • Energy not harnessed if reaction run in an open dish

    • All energy lost as heat to surroundings

  • Engineers seek to capture energy to do work

    • Maximize efficiency with which chemical energy is converted to work

    • Minimize amount of energy transformed to unproductive heat


Thermodynamically reversible

Thermodynamically Reversible

  • Process that can be reversed and is always very close to equilibrium

    • Change in quantities is infinitesimally small

  • Example - expansion of gas

    • Done reversibly, it does most work on surroundings


G maximum possible work

G° = Maximum Possible Work

  • G°is maximum amount of energy produced during a reaction that can theoretically be harnessed as work

    • Amount of work if reaction done under reversible conditions

    • Energy that need not be lost to surroundings as heat

    • Energy that is “free” or available to do work


Ex 5 calculate g

Ex. 5 Calculate G°

CalculateG° for reaction below at 1 atm and 25 °C, given H° = –246.1 kJ/mol, S° = 377.1 J/(mol K).

H2C2O4(s) + ½O2(g) → 2CO2(g) + H2O(l)

G°25 = H – TS

G° =(–246.1 – 112.4) kJ/mol

G° = –358.5 kJ/mol


Your turn7

Your Turn!

Calculate G°for the following reaction,

H2O2(l) → H2O(l) + O2(g)

given

H°= –196.8 kJ mol–1 and S°= +125.72 J K–1 mol–1.

A. –234.3 kJ mol–1

B. +234.3 kJ mol–1

C. 199.9 kJ mol–1

D. 3.7 × 105 kJ mol–1

G°= –196.8 kJ mol–1 – 298 K (0.12572 kJ K–1 mol-1)

G°= –234.3 kJ mol


G and position of equilibrium

G° and Position of Equilibrium

  • When G° > 0 (positive)

    • Position of equilibrium lies close to reactants

    • Little reaction occurs by the time equilibrium is reached

    • Reaction appears nonspontaneous

  • When G° < 0 (negative)

    • Position of equilibrium lies close to products

    • Mainly products exist by the time equilibrium is reached

    • Reaction appears spontaneous


G and position of equilibrium1

G° and Position of Equilibrium

  • When G° = 0

    • Position of equilibrium lies ~ halfway between products and reactants

    • Significant amount of both reactants and products present at time equilibrium is reached

    • Reaction appears spontaneous, whether start with reactants or products

  • Can Use G° to Determine Reaction Outcome

    • G° large and positive

      • No observable reaction occurs

    • G° large and negative

      • Reaction goes to completion


At equilibrium

At Equilibrium

  • G°=–RT lnK and K = e–G°/RT

  • Provides connection between G°and K

  • Can estimate K at various temperatures if G°is known

  • Can get G°if K is known

    Relationship between K and G°


Learning check2

Learning Check

Ex. 6Given that H° = –97.6 kJ/mol, S° = –122 J/(mol K), at 1 atm and 298 K, will the following reaction occur spontaneously?

MgO(s) + 2HCl(g)→ H2O(l) + MgCl2(s)

G° = H° – TS°

= –97.6 kJ/mol –298 K(–0.122 kJ/mol K)

G° = –97.6 kJ/mol + 36.4 kJ/mol

=–61.2 kJ/mol


Effect of change in pressure or concentration on g

Effect of Change in Pressure or Concentration on ∆G

  • Gat nonstandard conditions is related to G° at standard conditions by an expression that includes reaction quotient Q

  • This important expression allows for any concentration or pressure

  • Recall:


Ex 7 calculating g at nonstandard conditions

Ex. 7 Calculating G at Nonstandard Conditions

  • Calculate G at 298 K for the Haber process

    • N2(g) + 3H2(g) 2NH3(g)G° = –33.3 kJ

  • For a reaction mixture that consists of 1.0 atm N2, 3.0 atm H2 and 0.5 atm NH3

    Step 1 Calculate Q


Ex 7 calculating g at nonstandard conditions1

Ex. 7 Calculating G at Nonstandard conditions

Step 2 Calculate G= G°+ RT lnQ

G= –33.3 kJ/mol + (8.314 J/K mol)(1 kJ/1000J)(298K)(ln(9.3  10–3))

= –33.3 kJ/mol + (2.479 kJ/mol)(ln(9.3  10–3))

= –33.3 kJ + (–11.6 kJ/mol) = – 44.9 kJ/mol

  • At standard conditions all gases (N2, H2 and NH3) are at 1 atm of pressure

  • Gbecomes more negative when we go to 1.0 atm N2, 3.0 atm H2 and 0.5 atm NH3

  • Indicates larger driving force to form NH3

    • Preactants > Pproducts


Free energy diagrams

Free Energy Diagrams

Non-spont. G° > 0

  • G is different than G°

  • G interpreted as the slope of the free energy curve and tells which direction reaction proceeds to equilibrium

  • Minimum, at G = 0, indicates composition of equilibrium mixture

  • Because G ° is positive equilibrium lies closer to reactants

N2O4(g)  2NO2(g)

Equilibrium occurs here

at Ptotal = 1 atm with

16.6% N2O4 decomposed


Free energy diagrams1

Free Energy Diagrams

Spontaneous Rxn G°< 0

  • Reaction proceeds toward equilibrium from either A or B where slope, G, is not zero

  • The curve minimum, where G = 0, lies closer to the product side for spontaneous reactions. This is determined by G °


How k is related to g

How K is related to G°

  • Use relation G= G° + RT lnQ to derive relationship between K and G°

  • At Equilibrium

    ∆G = 0 andQ = K

  • So 0 = G°+ RT lnK

    G°=–RT lnK

    Taking antilog (ex) of both sides gives

    K = e–G°/RT


Ex 8 calculating g from k

Ex. 8 Calculating G° from K

  • Ksp for AgCl(s) at 25 °C is 1.8  10–10 Determine G° for the process

  • Ag+(aq) + Cl–(aq)  AgCl(s)

  • Reverse of Ksp equation, so

    G° =–RT lnK =

    –(8.3145 J/K mol)(298 K) × ln(5.6  109)(1 kJ/1000 J)

  • G°= –56 kJ/mol

  • Negative G°indicates precipitation will occur


System at equilibrium

System at Equilibrium

  • Neither spontaneous nor nonspontaneous

  • In state of dynamic equilibrium

  • Gproducts= Greactants

  • G = 0

  • Consider freezing of water at 0˚C

  • H2O(l)  H2O(s)

    • System remains at equilibrium as long as no heat added or removed

    • Both phases can exist together indefinitely

    • Below 0 °C, G < 0 freezing spontaneous

    • Above 0°C, G > 0 freezing nonspontaneous


No work done at equilibrium

No Work Done at Equilibrium

  • G = 0

  • No “free” energy available to do work

  • Consider fully charged battery

    • Initially

      • All reactants, no products

      • G large and negative

      • Lots of energy available to do work

    • As battery discharges

      • Reactants converted to products

      • ∆G less negative

      • Less energy available to do work

    • At equilibrium

      • G = Gproducts – Greactants = 0

      • No further work can be done

      • Dead battery


Phase change equilibrium

Phase Change = Equilibrium

  • H2O(l)  H2O(g)

  • G = 0 = H – TS

  • Only one temperature possible for phase change at equilibrium

    • Solid-liquid equilibrium

      • Melting/freezing temperature (point)

    • Liquid-vapor equilibrium

      • Boiling temperature (point)

  • Thus H = TS and

  • or


Ex 9 calculate t bp

Ex. 9 Calculate Tbp

CalculateTbp for reaction below at 1 atm and 25 °C, given H° = 31.0 kJ/mol, S° = 92.9 J/ mol K

Br2(l) → Br2(g)

For T > 334 K, G < 0 and reaction is spontaneous (S° dominates)

For T < 334 K, G > 0 and reaction is nonspontaneous (H° dominates)

For T = 334 K, G = 0 and T = normal boiling point


Learning check3

Learning Check

What is the expected melting point for Cu?

Cu(s)  Cu(l)

H = 1mol(341.1 kJ/mol – 1mol(0 kJ/mol)

H = +341.1kJ

S = 1mol(166.29 J/mol K – 1mol(33.1 J/mol K)

S = 133.19 J/K


Effect of temperature on g

Effect of Temperature on G°

  • Reactions often run at temperatures other that 298 K

  • Position of equilibrium can change as G° depends on temperature

    G° = H° – TS°

  • For temperatures near 298 K, expect only very small changes in H° and S°

  • For reaction at temperature,we can write:


Ex 10 determining effect of temperature on spontaneity

Ex. 10 Determining Effect of Temperature on Spontaneity

  • Calculate G° at 25 °C and 500 °C for the Haber process

    N2(g) + 3H2(g) 2NH3(g)

  • Assume that H° and S° do not change with temperature

  • Solving strategy

    Step 1. Using data in Tables 6.2 and 18.2 calculate H° and S° for the reaction at 25 °C

    • H° = – 92.38 kJ

    • S° = – 198.4 J/K


Ex 10 determining effect of temperature on spontaneity1

Ex. 10 Determining Effect of Temperature on Spontaneity

Step 2.Calculate G°for the reaction at 25 °C using H° and S°

N2(g) + 3H2(g)  2NH3(g)

  • H° = –92.38 kJ

  • S° = –198.4 J/K

  • G° =H° – TS°

  • G° = –92.38 kJ – (298 K)(–198.4 J/K)

  • G° = –92.38 kJ + 59.1 kJ = –33.3 kJ

  • So the reaction is spontaneous at 25 °C


  • Ex 10 determining effect of temperature on spontaneity2

    Ex. 10 Determining Effect of Temperature on Spontaneity

    Step 3.Calculate G°for the reaction at 500 °C using H° and S°.

    • T = 500 °C + 273 = 773 K

    • H° = – 92.38 kJ

    • S° = – 198.4 J/K

  • G°=H°– TS°

  • G°= –92.38 kJ – (773 K)(–198.4 J/K)

  • G°= –92.38 kJ + 153 kJ = 61 kJ

  • So the reaction is NOT spontaneous at 500 °C


  • Ex 10 does this answer make sense

    Ex. 10 Does this answer make sense?

    • G°=H°– TS°

      • H° = –92.38 kJ

      • S° = –198.4 J/K

    • Since both H° and S° are negative

    • At lowtemperature

      • G°will be negative and spontaneous

    • At high temperature

      • TS°will become a bigger positive number and

      • G°will become more positiveand thuseventually, at high enough temperature, will become nonspontaneous


    Ex 11 calculating k from g

    Ex. 11 Calculating Kfrom G°

    Calculate K at 25 °C for the Haber process

    N2(g) + 3H2(g) 2NH3(g)

    • G° = –33.3 kJ/mol = –33,300 J/mol

      Step 1 Solve for exponent

      Step 2 Take ex to obtain K

      Large K indicates NH3 favored at room temp.


    Your turn8

    Your Turn!

    Calculate the equilibrium constant for the

    decomposition of hydrogen peroxide at 298 K given

    G° = –234.3 kJ mol.

    A. 8.5 × 10-42

    B. 1.0 × 10499

    C. 3.4 × 10489

    D. 1.17 × 1041


    Ex 12 calculating k from g first calculate g

    Ex. 12 Calculating K from G°, First Calculate G°

    Calculate the equilibrium constant at 25 °C for the decarboxylation of liquid pyruvic acid to form gaseous acetaldehyde and CO2.


    Ex 12 first calculate g from g f

    Ex. 12 First Calculate ∆G° from ∆Gf°


    Ex 12 next calculate equilibrium constant

    Ex. 12 Next Calculate Equilibrium Constant

    K =1.85  1011


    Temperature dependence of k

    Temperature Dependence of K

    • G°=–RT lnK = H°– TS°

    • Rearranging gives

    • Equation for line

      • Slope = –H°/RT

      • Intercept = S°/R

    • Also way to determine K if you know H°and S°


    Ex 12 calculate k given h and s

    Ex. 12 Calculate K given H°and S°

    Calculate K at 500°C for Haber process

    N2(g) + 3H2(g) 2NH3(g)

    • Given H°= –92.38 kJ and S°= –198.4 J/K

  • Assume that H°andS°do not change with T

    ln K = + 14.37 – 23.86 = –9.49

    K = e–9.49 = 7.56 × 10–5


  • Bond energy

    Bond Energy

    • Amount of energy needed to break chemical bond into electrically neutral fragments

    • Useful to know

    • Within reaction

      • Bonds of reactants broken

      • New bonds formed as products appear

    • Bond breaking

      • First step in most reactions

      • One of the factors that determines reaction rate

      • e.g. N2 very unreactive due to strong NN bond


    Bond energies

    Bond Energies

    • Can be determined spectroscopically for simple diatomic molecules

      • H2, O2, Cl2

    • More complex molecules, calculate using thermochemical data and Hess’s Law

      • UseH°formation enthalpy of formation

    • Need to define new term

    • Enthalpy of atomization or atomization energy,Hatom

      • Energy required to rupture chemical bonds of one mole of gaseous molecules to give gaseous atoms


    Determining bond energies

    Determining Bond Energies

    • Ex. CH4(g) C(g) + 4H(g)

    • Hatom = energy needed to break all bonds in molecule

    • Hatom /4 = average bond C—H dissociation energy in methane

      • D = bond dissociation energy

        • Average bond energy to required to break all bonds in molecule

      • How do we calculate this?

        • Use H°f for forming gaseous atoms from elements in their standard states

        • Hess’s Law


    Determining bond energies1

    Determining Bond Energies

    • Path 1: Bottom

      • Formation of CH4 from its elements = H°f

    • Path 2: Top 3 step path

      • Step 1: break H—H bonds

      • Step 2: break C—C bonds

      • Step 3: form 4 C—H bonds

    • 2H2(g) 4H(g) H°1 = 4H°f (H,g)

    • C(s)  C(g)H°2 = H°f (C,g)

    • 4H(g) +C(g)  CH4(g) H°3 = –H°atom

      2H2(g) +C(s)  CH4(g) H° = H°f(CH4,g)


    Calculating h atom and bond energy

    Calculating Hatom and Bond Energy

    H°f(CH4,g) = 4H°f(H,g) + H°f(C,g) – Hatom

    • Rearranging gives

      Hatom= 4H°f(H,g) + H°f(C,g) – H°f(CH4,g)

    • Look these up in Table 18.3, 6.2 or appendix C

      Hatom= 4(217.9kJ/mol) + 716.7kJ/mol – (–74.8kJ/mol)

      Hatom= 1663.1 kJ/mol of CH4

    = 415.8 kJ/mol of C—H bonds


    Table 19 4 some bond energies

    Table 19.4 Some Bond Energies


    Using bond energies to estimate h f

    Using Bond Energies to Estimate H˚f

    • Calculate H˚f for CH3OH(g) (bottom reaction)

    • Use four step path

      • Step 1: break C—C bonds

      • Step 2: break H—H bonds

      • Step 3: break O—O bond

      • Step 4: form 4 C—H bonds


    Using bond energies

    Using Bond Energies

    H°f(CH3OH,g) =

    H˚f (C,g) + 4H˚f (H,g) +  H˚f (O,g) – Hatom (CH3OH,g)

    • H°f(C,g) + 4H˚f (H,g) + H˚f (O,g) = [716.7 +(4 × 217.9) + 249.2] kJ = +1837.5 kJ

    • Hatom (CH3OH,g) = 3DC—H + DC—O + DO—H= (3 × 412) + 360 + 463 = 2059 kJ

    • H°f(CH3OH,g) = +1837.5 kJ – 2059 kJ = –222 kJ

    • Experimentally find  H˚f (CH3OH,g) = –201 kJ/mol

    • So bond energies give estimate within 10% of actual


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