thermodynamics 11.ppt - bakersfieldcollege.edu b1b handouts fall 2012...enthalpy changes and...

2/7/2012

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Thermodynamics

1

• Study of energy changes and flow of energy

• Answers several fundamental questions:

– Is it possible for reaction to occur?

– Will reaction occur spontaneously (without outside

interference) at given T?

– Will reaction release or absorb heat?

• Tells us nothing about time frame of reaction

– Kinetics

• Two major considerations—must be balanced

– Enthalpy changes, ∆H (heats of reaction)

• Heat exchange between system and surroundings

– Nature's trend to randomness or disorder

Thermodynamics

2

Examples

of

Spontaneous Reactions

3

2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)

Consider this Reaction

Concerning this reaction:

1. Does this reaction naturally occur as written?

2. Will the reaction mixture contain sufficient

amount of product at equilibrium?

4

We can answer these questions with

heat measurements only!!!

1. We can predict the natural direction.

How?

2. We can determine the composition of the

mixture at equilibrium.

5

Consider the Laws of Thermodynamics

1st Law: The change in internal energy of a

system ∆U, equals q + w

∆E = ∆U = q + w

2nd Law: The total entropy of a system and its

surroundings increases for a spontaneous

process.

3rd Law: A substance that is perfectly crystalline

at 0 K has an entropy of zero.

You can’t win

You cant break even.

You can’t quit the game

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• Internal energy, E or U

– 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

systemsystemsystem PEKEE )()( +=

initialfinal EEE −−−−====∆∆∆∆

reactantsproducts EEE −−−−====∆∆∆∆

Review of First Law of Thermodynamics

7

q + Heat absorbed by system Esystem ↑

q −−−− Heat released by system Esystem ↓

w + Work done on system Esystem ↑

w −−−− Work done by system Esystem ↓

• Heat q =∆∆∆∆H Work w = P∆∆∆∆V

• ∆∆∆∆U = ∆∆∆∆E = q + w

• Conventions of heat and work

Two Methods of Energy Exchange Between

System and Surroundings

8

∆U(Internal Energy) is a state function:

Most often we are interested in the change:

Internal Energy = ∆Ε = ∆U = Uf - Ui

q = Energy that moves in and out of a system

w = Force x distance = F x d = P∆V

9

• Energy can neither be created nor destroyed

• It can only be converted from one form to another

– KE ↔ PE

– Chemical ↔ Electrical

– Electrical ↔ Mechanical

• E and ∆E are state functions

– Path independent

– ∆∆∆∆E = -∆∆∆∆U = q + w

First Law of Thermodynamics

systemsystemsystem PEKEE )()( +=10

VPq)VP(qE ∆∆∆∆−−−−====∆∆∆∆−−−−++++====∆∆∆∆

1. Electrical

2. Pressure-volume or P∆∆∆∆V

– w = −−−− P∆∆∆∆V

• Where P = external pressure

– If P∆∆∆∆V only work in chemical system, then

Work in Chemical Systems

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Heat vs. Work

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Showing work is P∆V

w = - F x h

= - F x ∆V/A

= -F/A x ∆V

= -P∆V

13

Reaction done at constant V

∆V = 0

P∆V = 0, so

∆E = qV

Entire E change due to heat absorbed or

lost

Heat at Constant Volume

14

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 pressure15

q = 165 J

P∆V = -92 J

∆U = q + w = (+165) + (-92) = +73 J16

Heat of Reaction and Internal Energy

Zn (s) + HCl (l) → ZnCl2 (aq) + H2 (g)

17

18

• Here the system expands and evolves

heat from A to B.

Zn2+(aq) + 2Cl-(aq) + H2(g)

∆V is positive, so work is negative.

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19

q = - w = -

Zn (s) + HCl (l) → ZnCl2 (aq) + H2 (g)

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w = -P∆V

= -(1.01 x 105 pa)(24.5 l)

= -1.01 x 105 pa)(24.5 x 10-3m3)

= -2.47 x 103 J

= -2.47 kJ

q = -152 kJ

∆U = -152 kJ + (-2.47 kJ) = -154.9 kJ

20

Diagram and explain the change in internal energy

for the following reaction.

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

q = -890.2 kJ

w = P∆V = +(1.01 x 105 pa)(24.5 l)(2)

= +(1.01 x 105 pa)(24.5 x 10-3m3)(2)

= +4.95 kJ

∆U = -890.2 kJ + (+4.95 kJ) = -885.2 kJ

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Enthalpy and Enthalpy Change

Enthalpy is defined as qp

Enthalpy = H = U + PV

All are state functions.

∆H = Hf - Hi

∆H = (Uf + PVf) – (Ui + PVi) = (Uf –Ui) + P(Vf –Vi)

∆U = qp - P∆V

∆H = (qp - P∆V) + P∆V = qp

∆Hof = Σn∆Ho

f (products) - Σm∆Hof (reactants)

∆Hof = Standard enthalpy change (25oC) 22

∆Hof = Σn∆Ho



Calculate ∆Hof for the reaction in slide 15

∆Hof = for NH3 (g) = -45.9 kJ

= for CO2 (g) = -393.5 kJ

= for NH2CONH2 = -319.2 kJ

= for H2O (l) = -285.8 kJ

∆Ho = [(-319.2 – 285.8) – (-2 x 45.9 – 393.50)] kJ = -119.7

Since ∆Ho has a negative sign, heat is evolved

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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

P

nRTV ====

∆∆∆∆====∆∆∆∆

P

nRTV

∆∆∆∆====∆∆∆∆

P

RTnV

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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

∆∆∆∆∗∗∗∗++++∆∆∆∆====∆∆∆∆

P

RTnPEH gas

RTnEH gas∆∆∆∆++++∆∆∆∆====∆∆∆∆

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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)

reactants)()( ooo

fproductsf HHH ∆∆∆∆−−−−∆∆∆∆====∆∆∆∆

)O(N2)(O)(NO4 5222oooo

fff HHHH ∆∆∆∆−−−−∆∆∆∆++++∆∆∆∆====∆∆∆∆

∆∆∆∆H° = 113 kJ 26

Step 2: Calculate

∆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)(8.31451 J/K·mol)(298 K)(1 kJ/1000 J)

∆E = 113 kJ – 7.43 kJ = 106 kJ

Ex. 1. (cont.)

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Ex. 1. finish

Step 4: Calculate % difference

Bigger than most, but still small

Note: Assumes that volumes of solids and liquids

are negligible

Vsolid ≈≈≈≈ Vliquid << Vgas

%.%kJ

kJ.66100

113

437difference % ====∗∗∗∗====

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Is Assumption that Vsolid ≈ Vliquid << Vgas Justified?

• Consider

CaCO3(s) + 2H+(aq) →→→→ Ca2+(aq) + H2O(l) + 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

HE ∆∆∆∆≈≈≈≈∆∆∆∆29

Learning Check• Consider the following reaction for picric acid:

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

• What type of reaction is it?

• Calculate ΔΗ°, ΔΕ°

8O2(g) + 2C6H

2(NO

2)

3OH(ℓ) → 3N2(g) + 12CO2(g) + 6H2O(ℓ)

ΔΗ°f

(kJ/mol) 0.00 3862.94 0.00 -393.5 -241.83

ΔΕ° = ΔH° – ΔngasRT = ΔH° – (15 – 8)mol*298K*

8.314×10–3kJ/(mol·K)

ΔH0 = 12mol(–393.5 kJ/mol) + 6mol(–241.83kJ/mol) +

6mol(0.00kJ/mol) – 8mol(0.00kJ/mol) – 2mol(3862.94kJ/mol)

ΔH0 = – 13,898.9 kJ

∆Ε° = –13,898.9 kJ – 29.0 kJ = – 13,927.9 kJ30

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Your Turn!Given the following:

3H2(g) + N2(g) → 2NH3(g) ∆Ho = -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

• ∆H = ∆E + ∆nRT ∆E = ∆H - ∆nRT

• ∆E =-46.19 kJ mol – (∆NRT)

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

• ∆E = -41.23 kJ mol-131

∆Hof = Σn∆Ho



Calculate ∆Hof for the reaction in slide 15

∆Hof = for NH3 (g) = -45.9 kJ

= for CO2 (g) = -393.5 kJ

= for NH2CONH2 = -319.2 kJ

= for H2O (l) = -285.8 kJ

∆Ho = [(-319.2 – 285.8) – (-2 x 45.9 – 393.50)] kJ = -119.7

Since ∆Ho has a negative sign, heat is evolved

32

Spontaneity and Entropy

Definition of Spontaneous Process:

Physical or chemical process that occurs by itself.

Give several spontaneous processes:

Still cannot predict spontaneity…..

Why??

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Enthalpy Changes and Spontaneity

• What are relationships among factors that influence

spontaneity?

• Spontaneous Change

– Occurs by itself

– Without outside assistance until

finished

• Ex.

– Water flowing over waterfall

– Melting of ice cubes in glass

on warm day

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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

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Entropy and the 2nd Law of Thermodynamics

2nd Law: The total entropy of a system and its

surroundings increases for a spontaneous

process.

Entropy = S = A thermodynamic quantity that is a

measure of how dispersed the energy

is among the different possible ways

that a system can contain energy.

Consider:

1. A hot cup of coffee on the table

2. Rock rolling down the side of a hill.

3. Gas expanding.

4. Stretching a rubber band.36

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Reaction Rate and Spontaneity

• ∆H indicates if reaction has tendency to occur

• Rate of reaction also plays role

– Some very rapid:

• Neurons firing in nerves in response to pain

• Detonation of stick of dynamite

– Some gradual:

• Erosion of stone

• Ice melting

• Iron rusting

– Some so slow, appear to be nonspontaneous:

• Gasoline and O2 at RT

• Many biochemical processes

Flask connected to an evacuated flask by a

valve or stopcock

∆S = Sf - Si

H2O (s) → H2O (l)

∆S = (63 – 41) J/K = 22 J/K

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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

Early thoughts were that exo were spontaneous)

39

Direction of Spontaneous Change

• Some endothermic reactions occur

spontaneously:

– Ice melting

– Evaporation of water from lake

– Expansion of CO2 gas into vacuum

• ∆H and ∆E are positive

– Heat absorbed

– Energy entering system

• Clearly other factors influence spontaneity

40

Concept Check: You have a sample of solid iodine

at room temperature. Later you notice

that the iodine has sublimed. What

can you say about the entropy change

of the iodine?

41

There are how many Laws of thermodynamics?

a. 1b. 2c. 3

d. 4e. 5

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2nd Law: The total entropy of a system and its surroundings increases for a spontaneousprocess.

Process occurs naturally as a result of energy dispersalIn the system.

∆S = entropy created + q/T

∆S > q/T

For a spontaneous process at a given temperature, thechange in entropy of the system is greater than the heat divided by the absolute temperature.

43

Entropy and Molecular Disorder

44

Reactants

Products

Energy

Reaction Progress

Thermodynamic vs. Kinetics• Thermodynamics tells us:

– Direction of reaction

– Is it possible for reaction to

occur?

– Will reaction occur

spontaneously at

given T?

– Will reaction release

or absorb heat?

– Kinetics tells us:

– Speed of reaction

– Pathway between reactants

and products

• Domain of

Thermodynamics

(initial and final states)

Domain of Kinetics

(reaction pathway)

45

Heat Transfer Between Hot and Cold Objects

• Consider system of two objects

– Initially one hot and one cold

– “Hot” = higher KEave of molecules = faster

– “Cold” = lower KEave of molecules = slower

• When they collide, what is most likely to occur?

– Faster objects bump into colder objects and transfer energy so…

– “Hot” objects cool down and slow down

– “Cold” objects warm up and speed up

– The reverse doesn’t occur

Heat Transfer Between Hot and Cold Objects

Result:

• Heat flows spontaneously from hot to colder object

• Heat flows because of probable outcome of

intermolecular collisions

• Spontaneous processes tend to proceed from states

of low probability to states of higher probability

• Spontaneous processes tend to disperse energy

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 random and ∴∴∴∴ more probable

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Entropy

• If Energy = money

• Entropy (S) describes number of different ways of

counting it

Criterion for Spontaneity• Clear order to things

– 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

• Measure of randomness and disorder

• Measure of chaos

• State function

• Independent of path

• ∆S = Change in Entropy

• Also state function

• For chemical reaction

initialfinal SSS −−−−====∆∆∆∆

reactantsSSS products −−−−====∆∆∆∆51

Entropy• Sproducts > Sreactants means ∆S +

– Entropy ↑'s

– Probability of state ↑'s

– Randomness ↑'s

– Favors spontaneity

• Sproducts < Sreactants means ∆S −−−−

– Entropy ↓'s

– Probability of state ↓'s

– Randomness ↓'s

– Does not favor spontaneity

• Any reaction that occurs with ↑ in entropy tends to occur spontaneously

52

Effect of Volume on Entropy

• For gases, Entropy ↑ as Volume ↑

A. Gas separated from vacuum by partition

B. Partition removed

C. Gas expands to achieve more probable particle

distribution

• More random, higher probability, more positive S53

Effect of Temperature on Entropy

• As T ↑, entropy ↑A. T = 0 K, particles (●) in equilibrium lattice positions and S

relatively low

B. T > 0 K, molecules vibrate, S ↑

C. T ↑ further, more violent vibrations occur and S higher than in B

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Effect of Physical State on Entropy

• Crystalline solid very low S

• Liquid higher S, molecules can move freely

– More ways to distribute KE among them

• Gas highest S, particles randomly distributed throughout container

– Many, many ways to distribute KE

55

Entropy Affected by Number of Particles

• Adding particles to system

• ↑ number of ways energy can be distributed in system

• So all other things being equal

• Reaction that produces more particles will have positive ∆S

56

Summary

• Larger V, greater ∆S

– Expansion of gas ∆S +

• Higher T, greater ∆S

– Higher T, means more KE in particles, move more, so random

distributions favored

• Ssolid < Sliquid << Sgas

– Solids more ordered than liquids, which are much more ordered

than gases

Reactions involving gases

• Simply calculate change in number of mole gas,

∆ngas

– If ∆ngas + , ∆S +

– If ∆ngas −−−− , ∆S −−−−57

Entropy Change for a Reaction

∆So may be positive for reactions if/with the following:

1. The reaction is one in which a molecule is

broken into two or more smaller molecules.

2. The reaction is one in which there is an increase

in moles of gas.

3. The process is one in which a solid changes to

a liquid or a liquid changes to a gas.

Did You Get It!

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

59

Entropy Changes in Chemical Reactions

Ex. N2 (g) + 3 H2(g) → 2 NH3(g)

nreactant = 4 nproduct = 2

∆n = 2 – 4 = –2

Predict ∆Srxn < 0

Higher positional

probability

Lower positional

probability

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Entropy Changes in Chemical Reactions

Reactions without gases

• Simply calculate number of mole molecules

∆n = nproducts – nreactants

– If ∆n + , ∆S +

– If ∆n −−−− , ∆S −−−−

– More molecules, means more disorder

– Usually the side with more molecules, has less complex

molecules

• Smaller, fewer atoms per molecule

Ex. 2 Predict Sign of ∆S for Following Reactions

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

– ∆ngas = 1 mol – 0 mol = 1 mol

– ∴ ∆ngas +, ∆S +

2 N2O5 (g) →→→→ 4 NO2 (g) + O2 (g)

– ∆ngas = 4 mol + 1 mol – 2 mol = 3 mol

– ∴ ∆ngas +, ∆S +

OH−−−− (aq) + H+ (aq) →→→→ H2O (l)

– ∆ngas = 0 mol

– ∆n = 1 mol – 2 mol = –1 mol

– ∴ ∆n −−−− , ∆S −−−−

Predict Sign of ∆S in 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)

positive

positive

positive

negative

CO2(s)→ CO2(g)

H2O(g) → H2O(ℓ)

negative

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

A. H2O(ℓ)

B. NaCl(aq)

C. AlCl3(s)

D. Can’t tell from the information

Did You Get It?

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)

64

Both Entropy and Enthalpy Can Affect Reaction

Spontaneity

• Sometimes they work together

– Building collapses

– PE ↓ ∆H −−−−

– Stones disordered ∆S +

• Sometimes work against each other

– Ice melting (ice/water mix)

– Endothermic

• ∆H + nonspontaneous

– ↑ Disorder of molecules

• ∆S + spontaneous

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

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Second Law of Thermodynamics

• When a spontaneous event occurs, total entropy of

universe ↑’s

– (∆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

T

qS

gssurroundingssurroundin =∆

2nd Law: The total entropy of a system and its surroundings increases for a spontaneousprocess.

Process occurs naturally as a result of energy dispersalIn the system.

∆S = entropy created + q/T

∆S > q/T

For a spontaneous process at a given temperature, thechange in entropy of the system is greater than the heat divided by the absolute temperature.

68

Does This Make Sense?

• Yes?

• As heat added

– Disorder or entropy ↑, ∴ S ∝ q

• But how much ↑ T, ↑ S, depends on T at

which it occurs

– Low T, larger ↑ S

– High T, smaller ↑ S

• ∴ S ∝ 1/T

What do you think?

When ice melts in your hand (assume your hand is 30o C),

A. the entropy change of the system is less then the entropy

change of the surroundings.

B.the entropy change of the surroundings is less than the

entropy change of the system.

C.the entropy change of the system equals the entropy

change of the surroundings.

• ∆Ssys = ∆H/273K ∆Ssurr = ∆H/303K

• ∆Ssys > ∆Ssurr

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

T

H

T

qS

systemgssurroundingssurroundin

∆−==∆

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 (+)

So,

(∆∆∆∆Hsystem – T∆∆∆∆Ssystem) < 0

– (–) for reaction to be spontaneous

T

HSS

systemsystemtotal

∆−∆=∆

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Standard Entropy (Absolute Entropy): Entropy

value for the standard state of a species.

See Table B-13 p. B-17 and Table B-16 p. B-28

Entropy values of substances must be positive.

So must be >0 but Ho can be plus or minus

(Why?)

How about ionic species?

So for H3O+ is set at zero.

73

Predict The Entropy sign for the following reactions:

a. C6H12O11 (s) → 2CO2 (g)+ C2H5OH (l)

b. 2 NH3 (g) + CO2 (g) → NH2CONH2 (aq) + H2O (l)

c. CO (g) + H2O (g) → CO2 (g) + H2 (g)

Exercise 13.2 p 578

d. Stretching a rubber band

74

Calculating ∆So for a reaction

∆So = Σn∆So (products) - Σm∆So (reactants)

Calculate the entropy change for the following reaction

At 25 oC.


So 2 x 193 214 174 70

∆So = Σn∆So (products) - Σm∆So (reactants)

∆So= [(174 + 70) - (2x 193 + 214)] J/K= -356 J/K

See exercise 13.3 p 584 and problems 8-12 and 23-2675

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)76

Consequences of Third Law

1. All substances have positive entropies as they are more

disordered than at 0 K

– Heating ↑↑↑↑ randomness

– S°is biggest for gases—most disordered

2. 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

∑∑ −=∆ )reactants()products( ooo SmSnS77

Learning Check

Calculate ∆S0 for the following:

• CO2(s) → CO2(g)

187.6 213.7 S0 (J/mol·K)

•CaCO3(s) → CO2(g) + CaO(s)

92.9 213.7 40 S0 (J/mol·K)

• ∆S0 = (213.7 – 187.6) J/mol·K

• ∆S0 = 26.1 J/mol·K

•∆S0 = (213.7 +40 – 92.9) J/mol·K

•∆S0 = 161 J/mol·K78

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14

Ex. 3. Calculate ∆∆∆∆S°for reduction of aluminum oxide by

hydrogen gas

Al2O3 (s) + 3 H2 (g) →→→→ 2 Al (s) + 3 H2O (g)

Substance S° (J/ K·mol)

Al (s) 28.3

Al2O3 (s) 51.00

H2 (g) 130.6

H2O (g) 188.7

∑∑ −=∆ ooo

reactantsSnSnS rproductsp

]3[

]32[

)(2)(32

)(2)(

oo

ooo

gs

gs

HOAl

OHAl

SS

SSS

+−

+=∆

79

Ex. 3

⋅+

⋅−

⋅+

⋅=∆

Kmol

130.6J*mol 3

Kmol

J51.00*mol 1

Kmol

J188.7*mol 3

Kmol

J28.3*mol 2oS

∆S°= 56.5 J/K + 566.1 J/K – 51.00 J/K – 391.8 J/K

∆S°= 179.9 J/K

80

Gibbs Free Energy

• Would like one quantity that includes all

three factors that affect spontaneity of

a reaction

• Define new state function

• 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

81

Gibbs Free Energy

∆∆∆∆G < 0 Spontaneous process

∆∆∆∆G = 0 At equilibrium

∆∆∆∆G > 0 Nonspontaneous

∆∆∆∆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

82

Criteria for Spontaneity?

• At constant P and T, process spontaneous only if it is

accompanied by ↓↓↓↓ in free energy of system

∆∆∆∆H ∆∆∆∆S Spontaneous?

– + ∆∆∆∆G = (–) – [T(+)] = – Always, regardless of T

+ – ∆∆∆∆G = (+) – [T(–)] = + Never, regardless of T

+ + ∆∆∆∆G = (+) – [T(+)] = ? Depends; spontaneous at

high T, –∆∆∆∆G

– – ∆∆∆∆G = (–) – [T(–)] = ? Depends; spontaneous at

low T, –∆∆∆∆G

83

Summary

• When ∆∆∆∆H and ∆∆∆∆S have

same sign, T determines

whether spontaneous or

non-spontaneous

• Temperature-controlled

reactions are spontaneous

at one temperature and

not at another

84

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15

Free Energy Concept

∆Ho – T∆So Can Serve as a criteria for Spontaneity


∆Ho = -119.7 kJ ∆So = -365 J/K = -0.365 kJ/K

∆Ho – T∆So = (-119.7 kJ) – (298 K) x (-0.365kJ/K

= -13.6 kJ

∆Ho – T∆So is a negativity quantity, from which we can

conclude that the reaction is spontaneous

under standard conditions.

G = H – T S

85

Free Energy and Spontaneity

Free Energy: Thermodynamic quantity defined by

the equation G = H - TS

∆G = ∆H - T∆S

If you can show that ∆G for a reaction at a given

temperature and pressure is negative, you can

predict that the reaction will be spontaneous…

86

Standard Free Energy Changes

Standard Conditions:

1 atm pressure

1 atm partial pressure

1 M concentration

Temperature of 25 oC or 298 K

Standard free energy is free-energy change that takes

place when reactants in their standard states are

converted to products in their standard states.

∆Go = ∆Ho - T∆So

87

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°– (298.15 K)∆∆∆∆S°

Ex. 4. Calculate ∆∆∆∆G°for reduction of aluminum oxide by

hydrogen gas

Al2O3(s) + 3 H2(g) →→→→ 2 Al(s) + 3H2O(g)

88

Ex. 4. Method 1

Al2O3 (s) + 3 H2 (g) →→→→ 2 Al (s) + 3 H2O (g)

Step 1: Calculate ∆∆∆∆H°for reaction using o

fH∆∆∆∆Substance (kJ/mol)

Al (s) 0.0

Al2O3 (s) –1669.8

H2 (g) 0.0

H2O (g) –241.8

reactantsHnproductsHnH frfp ∑∑ ∆−∆=∆ ooo

]3[

]32[

)(Hf)(OAlf

)O(Hf)Al(f

232

2

gs

gs

HH

HHH

oo

ooo

∆+∆−

∆+∆=∆

o

fH∆∆∆∆

89

Ex. 4. Method 1 Step 1 (∆H°)

++++

−−−−−−−−

−−−−++++

====

mol

kJ. mol*

mol

kJ. mol*

mol

kJ. mol*

mol

kJ. mol*∆H

0003

816691

82413

0002o

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

∆H°= 944.4 kJ

90

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Ex. 4 Method 1

Step 2: Calculate ∆S° see Ex. 4

∆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°= +

– ∴ not spontaneous

91

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

o

fG∆∆∆∆

reactantsGnproductsGnG frfp ∑∑ ∆−∆=∆ ooo

92

Ex. 4. Method 2

Calculate ∆∆∆∆G°for reduction of aluminum oxide

by hydrogen gas.

Al2O3 (s) + 3 H2 (g) →→→→ 2 Al (s) + 3 H2O (g)

]3[

]32[

)(Hf)(OAlf

)O(Hf)Al(f

232

2

gs

gs

GG

GGG

oo

ooo

∆+∆−

∆+∆=∆

Substance (kJ/mol)

Al (s) 0.0

Al2O3 (s) –1576.4

H2 (g) 0.0

H2O (g) –228.6

o

fG∆∆∆∆

93

Ex. 4. Method 2

+

−−

−+

=

mol

kJ. mol*

mol

kJ. mol*

mol

kJ. mol*

mol

kJ. mol*∆G

0003

415761

62283

0002o

∆G°= 0.0 kJ – 685.8 kJ – 0.00 kJ – (– 1576.4 kJ)

∆G°= 890.6 kJ

Both methods same

within experimental error94

∆Go as a Criterion for Spontaneity

∆Go = ∆Ho - T∆So

1. When ∆Go is a large negative number (more

negative than about – 10 kJ), the reaction is spontaneous

as written, and reactants transform almost entirely into

products when equilibrium is reached.

2. When ∆Go is a large positive number (more

positive than about + 10 kJ), the reaction is not

spontaneous as written, and reactants transform almost

entirely into products when equilibrium is reached.

3. When ∆Go has a small positive or negative value(less

than about 10 kJ), the reaction mixture gives an

equilibrium mixture with significant amounts of both

reactants and products.. 95

Interpreting the Sign of ∆Go

Calculate ∆Ho and ∆Go for the following reaction

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

Interpret the signs of ∆Ho and ∆Go

∆Hfo are as follows: KClO3 (s) = -397.7 kJ/mol

KCl (s) = -436.7 kJ/mol

O2 (g) = 0

∆Gfo are as follows: KClO3 (s) = -296.3 kJ/mol

KCl (s) = -408.8 kJ/mol

O2 (g) = 0

96

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17

2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

∆Hfo 2 x (-397.70) 2 x (-436.7) 0 kJ

∆Gfo 2 x (-296.3) 2 x (-408.8) 0 kJ

Then:

∆Ho = [2 x (-436.7) – 2 x (-397.7)] kJ = -78 kJ

∆Go = [2 x (-408.8) – 2 x (-296.3)] kJ = -225 kJ

The reaction is exothermic, liberating 78 kJ of heat. The

large negative value of ∆Go indicates that the equilibrium

is mostly KCl and O2.

97

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

98

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

99

∆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

100

Ex. 5

Calculate ΔG° for reaction below at 1 atm and

25°C, given ΔH° = –246.1kJ/mol, ΔS° = 377.1

/(mol·K).

H2C2O4(s) + ½ O2(g) → 2CO2(g) + H2O(ℓ)

∆∆∆∆G25 = ∆∆∆∆H – T∆∆∆∆S

ΔG° =(–246.1 – 112.4)kJ/mol

ΔG° = –358.5kJ/mol

⋅−−=∆

kJ

J

molK

JK

mol

kJG

1000

11.377)298(1.246o

101

You Try!Calculate ∆Go for the following reaction,

H2O2(l ) → H2O(l ) + O2(g)

given ∆Ho = -196.8 kJ mol-1 and ∆So = +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 x 105 kJ mol-1

•∆Go = -196.8 kJ mol-1 – 298 K (0.12572 kJ K-1 mol-1)

•∆Go = -234.3 kJ mol

102

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System at Equilibrium

• Neither spontaneous nor nonspontaneous

• In state of dynamic equilibrium

• Gproducts= Greactants

• ∆G = 0

• Consider freezing of water at 0o C

• H2O(ℓ) � H2O(s)

– System remains at equilibrium as long as no heat added

or removed

– Both phases can exist together indefinitely

– Below 0oC, ∆G < 0 freezing spontaneous

– Above 0oC, ∆G > 0 freezing nonspontaneous

•Define equilibrium103

Ex. 6

Calculate Tbp for reaction below at 1 atm and 25°C, given

ΔH° = 31.0kJ/mol, ΔS° = 92.9 J/(mol·K)

Br2(l) → Br2(g)

• For T > 334 K, ΔG < 0 and reaction is

spontaneous (ΔS° dominates)

• For T < 334 K, ΔG > 0 and reaction is

nonspontaneous (ΔH° dominates)

• For T = 334 K, ΔG = 0 and T = normal

boiling point

KKmolkJ

molkJ

S

HTbp 334

)/(0929.0

/0.31=

⋅≈

∆

∆≈

o

o

104

Free energy change during reaction

105

Free energy change during reaction

106

Free Energy and Equilibrium Constant

Very important relation is the relation between

free energy and the equilibrium constant.

Thermodynamic Equilibrium Constant- the equilibrium

constant in which the concentration of gases are

expressed in partial pressures in atmospheres,

whereas the concentration of solutes in liquid

are expressed in molarities.

K = Kc for reactions involving only liquid solutions

K = Kp for reactions involving only gases

107

Equilibrium Expression

Kc =

aA + bB ⇔ cC + dD

[C]c[D]d

[A]a[B]b

Solids and liquids considered unity (1)

108

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19

Write Kp and Kc

• Consider again

N2(g) + 3H2(g) ⇄ 2 NH3(g)

Kc = [NH3]

2

[N2][H2]3

109

• In terms of partial pressures,

N2(g) + 3H2(g) ⇄ 2 NH3(g)

Kp = [PNH3]

2

[PN2][PH2]3

110

• Kc and Kp are related.

• PV = nRT

• P = [n/V]RT

• Kp=Kc × (RT)∆n

• ∆n = (number of moles of product gas) –(number of moles of reactant gas)

• For this reaction, ∆n = -2.

N2(g) + 3H2(g) ⇔ 2 NH3(g)

111

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 battery112

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

• orT

HS

∆=∆

S

HT

∆

∆=

113

114

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20

∆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

115

∆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

116

Learning Check

Ex. 7 Given that ΔH° = –97.6 kJ/mol,

ΔS° = –122 J/(mol·K), at 1atm and 298K,

will the following reaction occur spontaneously?

• MgO(s) + 2HCl(g) → H2O(ℓ) + MgCl2(s)

∆G° = ∆H° – T∆S°

= –97.6kJ/mol – 298K(–0.122 kJ/mol⋅K)

∆G° = –97.6kJ/mol +36.4kJ/mol

= –61.2 kJ/mol

117

Effect of Temperature on ∆G°

• Reactions often run at T’s other that 298 K

• Position of equilibrium can change as ∆G°depends on T

– ∆G°= ∆H°−−−− T∆S°

• For T’s near 298 K, expect only very small changes in ∆H

and ∆S°

• For reaction at T, we can write:

ooo

TTT STHG ∆−∆=∆

ooo

298298 STHGT ∆−∆≈∆

118

Ex. 8 Determining Effect of T on Spontaneity

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

process

N2 (g) + 3 H2 (g) �� 2 NH3 (g)

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

• Solving Strategy

Step 1. Using data from Tables, calculate ∆∆∆∆H°and ∆∆∆∆S°

for the reaction at 25°C

– ∆∆∆∆H°= – 92.38 kJ

– ∆∆∆∆S°= – 198.4 J/K

119


Step 2. Calculate ∆∆∆∆G°for the reaction at 25°C using

∆∆∆∆H°and ∆∆∆∆S°

N2 (g) + 3 H2 (g) �� 2 NH3 (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

J

kJ

1000

1

120

2/7/2012

21


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

J

kJ

1000

1

121

Ex. 8 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 low T

– ∆∆∆∆G°will be negative and spontaneous

• At high T

– T∆∆∆∆S°will become a bigger positive number and

– ∆∆∆∆G°will become more positive and thus eventually, at

high enough T, will become nonspontaneous

122

Effect of Change in Pressure or Concentration on ∆G

• ∆G at 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:

QRTGG ln+∆=∆ o

x

y

Q][

][

reactants

products=

123

Ex. 9 Calculating ∆G at Nonstandard conditions

• Calculate ∆∆∆∆G at 298 K for the Haber process

– N2 (g) + 3 H2 (g) �� 2 NH3 (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

3

3

2

3

2

103.9)0.3)(0.1(

)50.0(

22

3 −×===HN

NH

PP

PQ

124

Ex. 9 Calculating ∆G at Nonstandard conditions

Step 2 Calculate ∆∆∆∆G = ∆G°+ RT lnQ

∆∆∆∆G = –33.3 kJ/mol + (8.314J/K·mol)*(1kJ/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

• ∆∆∆∆G becomes 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

125

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

126

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22

At Equilibrium• ∆G°= −−−−RT lnK and K = e−−−−∆G°/RT

• Provides connection between ∆G°and K

• Can estimate K’s at various T’s if know ∆G°

• Can get ∆G°in know K’s

Relationship between K and ∆G

Keq ∆∆∆∆G°°°° Reaction

> 1 − Spontaneous Favored

Energy released

< 1 + non-spontaneous Unfavorable

Energy needed

= 1 0 At Equilibrium127

Ex. 10 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 = –(8.3145J/K·mol)(298K)*ln(5.6 × 109)*(1kJ/1000J)

• ∆G°= –56 kJ/mol

• Negative ∆G°indicates precipitation will occur

9

10106.5

108.1

11×=

×==

−spK

K

128

Ex. 11 Calculating K from ∆G°

• Calculate K at 25°C for the Haber process

N2 (g) + 3 H2 (g) �� 2 NH3 (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 RT

54.13/ 107 ×=== ∆− eeK RTG o

4.13)298)(/3145.8(

)/300,33(=

⋅

−−=

∆−

KmolKJ

molJ

RT

G o

RTGeK /o∆−=

129

You Try!Calculate the equilibrium constant for the

decomposition of hydrogen peroxide at 298 K given

∆Go = -234.3 kJ mol.

A. 8.5 x 10-42

B. 1.0 x 10499

C. 3.4 x 10489

D. 1.17 x 1041

94.56 41

( 234,300 / )94.56

(8.3145 / )(298 )

1.17 x 10

G J mol

RT J K mol K

K e

−∆ − −= =

⋅

= =

o

130

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

H3CC

C

O

OH

O

H3CC

O

H+ CO2

131

First Calculate ∆G°from ∆Gf°

Compound ∆Gf°, kJ/mol

CH3COH −133.30

CH3COCOOH −463.38

CO2 −394.36

)()()( 323 COCOOHCHfCOfCOHCHf GGGG oooo ∆−∆+∆=∆

)38.463()36.394(30.133 −−−+−=∆ oG

kJG 28.64−=∆ o

132

2/7/2012

23

Next Calculate Equilibrium Constant

RTGeKo∆−=

94.251000

)298)(/314.8(

28.64−=×

−=

∆

kJ

J

KKJ

kJ

RT

Go

945.25)945.25( eeK == −−

K = 1.85 ×××× 1011

133

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°

R

∆S

T

1

R

∆Hln

oo

+

−=K

134

Ex. 12 Calculate K given ∆H° and ∆S°

• Calculate K at 500 °C for Haber process

N2 (g) + 3 H2 (g) �� 2 NH3 (g)

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

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

R

S

RT

HK

oo ∆+

∆−=ln

)/314.8(

/4.198

)773)(/314.8(

380,92ln

KJ

KJ

KKJ

JK

−+

−−=

lnK = + 14.37 – 23.86 = – 9.49

K = e–9.49 = 7.56 × 10–5

135

Calculation of ∆∆∆∆Go at Various Temperatures

Consider the following reaction:

CaCO3 (s) → CaO (s) + CO2 (g)

At 25 oC ∆Go = +130.9 and Kp = 1.1 x 10-23 atm

What do these values tell you about CaCO3?

What happens when the reaction is carried out at a

higher temperature?

136

Calculating ∆∆∆∆Go and K at Various Temperatures

a. What is ∆Go at 1000oC for the calcium carbonate reaction?


Is this reaction spontaneous at 1000oC and 1 atm?

b. What is the value of Kp at 1000oC for this reaction?

What is the partial pressure of CO2?

137

Strategy for solution…

a. Calculate ∆Ho and ∆So at 25 oC using standard

enthalpies of formation and standard entropies.

Then substitute into the equation for ∆Gfo.

b. Use the ∆Gfo value to find K (=Kp ) as in

Exercise 13.16.

138

2/7/2012

24

a. From Table B-16 you have the following:


∆Hfo: -1206.9 -635.1 -393.5 kJ

So: 92.9 38.2 213.7 J/K

∆Ho = [(-635.1 – 393.5) – (-1206.9)] = 178.3 kJ

∆So = [(38.2 + 213.7) – (92.9)] = 159.0 J/K

∆GTo = ∆Ho –T∆So

= 178.3kJ – (1273 K)(0.1590kJ/K) = -24.1 kJ

∆Go is negative – reaction is spontaneous 139

b. Substitute the values of ∆Go at 1273 K, which

equals -24.1 x 103 J, into the equation relating

ln K and ∆Go.

ln K = ∆Go

-RT=

-24.1 x 103

-8.31 x 1273= 2.278

K = Kp = e2.278 = 9.76

Kp = PCO2 = 9.76 atm

140

Where does the reaction change from spontaneous

to non-spontaneous?

∆Go = 0 = ∆Ho - T∆So

Solve for T;

T = ∆Ho

∆So =178.3 kJ

0.1590 kJ/K

T = 1121 K = 848 oC

141

142

• 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

– 1st step in most reactions

– One of the factors that determines reaction rate

– Ex. N2 very unreactive due to strong N ≡ N bond

Bond Energy

143

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 1 mole of gaseous molecules to give gaseous atoms

144

Determining Bond Energies

• Ex. CH4(g) →→→→ C(g) + 4 H(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

2/7/2012

25

145

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

1.2H2(g) → 4H(g) ∆H°1 = 4∆H°f (H,g)

2.C(s) → C(g) ∆H°2 = ∆H°f (C,g)

3.4H(g) + C(g) → CH4(g) ∆H°3 = –∆Hatom

2H2(g) + C(s) → CH4(g) ∆H°= ∆H°f(CH4,g)

146

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

4

kJ/mol 1663.1energy bond =

= 415.8 kJ/mol of C—H bonds

147

147

Table 19.4 Some Bond Energies

148

Using Bond Energies to Estimate ∆H°f

• Calculate ∆H°f for CH3OH(g) (bottom reaction)

• Use 4 step path

– Step 1 break 1C—C bonds

– Step 2 break 2H—H bonds

– Step 3: break 1O—O bond

– Step 4: form 3 C—H, 1 O—H, & 1O—C bonds

149

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

Chapter 13/19 Chemical Thermodynamics

1. Spontaneous Chemical and Physical Processes

2. Entropy and Disorder

3. Entropy and the Second Law of Thermodynamics

4. Standard-State Entropies of Reaction

5. The Third Law of Thermodynamics

6. Calculating Entropy Changes for Chemical Reactions

7. Gibbs Free Energy

8. The Effect of Temperature on the Free Energy of a Reaction

9. Beware of Oversimplification

10. Stand-State Free Energies of Reaction

11. Equilibria Expressed in Partial Pressures

12. Interpreting Stand-State Free Energy of Reaction Data

13. Relationship Between Free Energy and Equilibrium Constants

14. Temperature Dependence of Equilibrium Constants

15. Gibbs Free Energies of Formation and Absolute Entropies

16. Calculate ∆H with bond energies150

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Entropy Change for a Phase Transition

∆S > q/T (at equilibrium)

What processes can occur under phase change at equilibrium?

Solid to liquidLiquid to gas

Solid to gas

151

Solution:

∆S = ∆Hvap/T = (39.4 x 103 J/mol)/ 298 K= 132 j/(mol·K)

Entropy of Vapor = (216 + 132) J/(mol·K)= 348 J/(mol·K)

152

153

thermodynamics 11.ppt - bakersfieldcollege.edu b1b handouts fall 2012...enthalpy changes and...

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