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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

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
  • Heat qWork w
  • 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
slide11

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 mol T = 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 = 4 nproduct = 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°: 187.6 213.7 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
slide52
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 and Q = 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.

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

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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