thermodynamics and equilibrium. 2 –from chapter 6: the change in enthalpy equals the heat of...

Thermodynamics

and Equilibrium

2

– From Chapter 6: the change in enthalpy equals the heat of reaction at constant pressure

– In this chapter we will define enthalpy in terms of the energy of the system.

Thermodynamics• Thermodynamics the study of the

relationship between heat and other forms of energy in a chemical or physical process.

• Describes systems whose states are determined by thermal parameters.

3

– The internal energy, U, is the sum of the kinetic and potential energies of the particles making up the system.

U = Ek + Ep

– Internal energy is a state function - a property of a system that depends only on its present state.

First Law of Thermodynamics

4

– 1 mol of water at 0 oC and 1 atm pressure has a definite quantity of energy.

– When a system changes [environmental conditions change] from one state to another, its internal energy changes.

ΔU = Uproducts - Ureactants

initialfinal UUU


5

– Changes in U manifest themselves as exchanges of energy between the system and surroundings.

– These exchanges of energy are of two kinds; heat (q) and work (w).

– ΔU = q + w


6

– Heat is energy that moves into or out of a system because of a temperature difference between system and surroundings.

– Work is the energy exchange that results when a force F moves an object through a distance d; work (w) = Fd


7

– Obeying our sign convention.

Work done by the system is negative.

Work done on the system is positive.

Heat evolved by the system is negative.

Heat absorbed by the system is positive.


8

Heat and Work• What happens as heat is

absorbed?

Heat

Work

Weight

Piston

Gas filled chamber

–The system gains internal energy from the heat absorbed and loses internal energy via the work done.

9

– The first law of thermodynamics states that the change in internal energy, U, equals heat plus work.

wqU


10

– When gases are produced, they exert force on the surroundings as pressure.

– If the reaction is run at constant pressure, then the gases produced represent a change in volume analogous to a distance over which a force is exerted.

Heat of Reaction and Internal Energy• When a reaction is run in an

open vessel (at constant P), any gases produced represent a potential source of “expansion” work.

11

– You can calculate the work done by a chemical reaction simply by multiplying the atmospheric pressure by the change in volume, V.

– It follows therefore, that

– (force x distance)

VPw

Heat of Reaction and Internal Energy

12

– For example, when 1.00 mol Zn reacts with excess HCl, 1.00 mol H2 is produced.

– At 25 oC and 1.00 atm (1.01 x 105 Pa), this amount of H2 occupies 24.5 L (24.5 x 10-3 m3).

– The work done by the chemical system in pushing back the atmosphere is

)g(H)aq(Zn)aq(OH2)s(Zn 22

3 qp = -152.4 kJ


)m105.24()Pa1001.1(VPw 335 kJ 2.47-or J1047.2 3

13

W = -PΔV

• Therefore, since PV = nRT– -PΔV = -ΔnRT

– This indicates that work is done only if there is a change in the number of moles of gaseous reactants and products.

14

– Relating the change in internal energy to the heat of reaction, you have

wqU p

)kJ 47.2()kJ 4.152(U kJ 9.154U


15

– When the reaction occurs, the internal energy changes as the kinetic and potential energies change in going from reactants to products.

– Energy leaves the system mostly as heat (qp = -152.4 kJ) but partly as expansion work (-2.47 kJ).


16

Determine if work is done, and its sign in the following rxn…

SiO2(s) + 3C(s) → SiC(s) + 2CO(g)

Δn = # mol gaseous product - # moles gaseous reactants Δn = 2 – 0 Δn = 2 w= -PΔV=- Δn RT

• The sign of work is the opposite of Δn – Work is done [b/c difference in gaseous moles]– Work is done by the system

• b/c the sign of Δn is positive, so the sign of w is negative

17

• A 2.00 g sample of hydrazine, N2H4, is combusted in a bomb calorimeter containing 6400 g of water. The temperature increases 1.17oC. The heat capacity of the calorimeter is 3.76 kJ/oC.

• What is ΔU for hydrazine?N2H4 + O2 N2 + 2H2O

ΔU = Uproducts – Ureactants

– The energy changes in this situation are recorded in terms of heat and temperature.

18

• The heat causing the water temperature change…

qH2O = mΔTCp

qH2O = (6400 g)(1.17oC)(4.184 J/goC)

qH2O = 31,329 J

qH2O = 31.329 kJ• Heat absorbed by calorimeter…

3.76kJ x 1.17oC = 4.39 kJ oC• Total heat = 31.329 kJ + 4.39kJ = 35.72

kJ• ΔU = -35.72 kJ/ 2gN2H4

• Exothermic rxn

19

– We now define enthalpy, H , precisely as the quantity U + PV.

H = U + PV–allows pressure to be variable

– H is a state function.

Enthalpy and Enthalpy Change• In Chapter 6, we tentatively

defined enthalpy in terms of the relationship of H to the heat at constant pressure.

20

Oh so many equalities!

• ΔU = q + w• w = -PΔV = - ΔnRT• ΔU = q –PΔV• H = U + PV• ΔH = Hf – Hi, or (U+PV)f - (U+PV)I

• Δ H = (Uf – Ui) + P(Vf –Vi) *, or• ΔH = ΔU+P ΔV *, or• ΔH = (q – PΔV)+P ΔV *, or• ΔH = q *

*constant pressure

21

– This means that at a given temperature and pressure, a given amount of a substance has a definite enthalpy.

– Therefore, if you know the enthalpies of substances, you can calculate the change in enthalpy, H , for a reaction.

Enthalpy and Enthalpy Change• In Chapter 6, we tentatively

defined enthalpy in terms of the relationship of H to the heat at constant pressure.

22

– In practice, we measure certain heats of reactions and use them to tabulate enthalpies of formation, Ho

f.

– Standard enthalpies of formation for selected compounds are listed in Table 6.2 and in Appendix C.

Enthalpy and Enthalpy Change

23

– The standard enthalpy change for a reaction is

)reactants(Hm)products(HnH of

of

o

Enthalpy and Enthalpy Change

24

– Examples include:A rock at the top of a hill rolls down.

Heat flows from a hot object to a cold one.

An iron object rusts in moist air.

– These processes occur without requiring an outside force and continue until equilibrium is reached. (see Figure 19.4)

Spontaneous Processes and Entropy• A spontaneous process is a physical or

chemical change that occurs by itself.

25

– Entropy, S , is a thermodynamic quantity that is a measure of the randomness or disorder of a system.

– The SI unit of entropy is joules per Kelvin (J/K) and, like enthalpy, is a state function.

Entropy and the Second Law of Thermodynamics

26

– The net change in entropy of the system, S , equals the sum of the entropy created during the spontaneous process and the change in energy associated with the heat flow.

Entropy and the Second Law of Thermodynamics• The second law of

thermodynamics states that the total entropy of a system and its surroundings always increases for a spontaneous process.

27

– In other words,

Tq

created entropy S

(For a spontaneous process)


28

– Normally, the quantity of entropy created during a spontaneous process cannot be directly measured.

– If we delete it from our equation, we can conclude


Tq

S

(For a spontaneous process)

29

– The restatement of the second law is as follows:

– For a spontaneous process, the change in entropy of the system is greater than the heat divided by the absolute temperature.

– Under these conditions, no significant amount of entropy is created.

– The entropy results entirely from the absorption of heat. Therefore,


Tq

S (For an equilibrium process)

30

– Other phase changes, such as vaporization of a liquid, also occur under equilibrium conditions.

– Therefore, you can use the equation

to obtain the entropy change for a phase change.

Entropy Change for a Phase Transition• If during a phase transition,

such as ice melting, heat is slowly absorbed by the system, it remains near equilibrium as the ice melts.

Tq

S

31

– When liquid CCl4 evaporates, it absorbs heat: Hvap = 43.0 kJ/mol (43.0 x 103 J/mol) at 25 oC, or 298 K. The entropy change, S, is

– Remember H = q

)KJ/(mol 144K 298

mol/J100.43T

HS

3vap

A Problem To Consider

• The heat of vaporization, Hvap of carbon tetrachloride, CCl4, at 25 oC is 43.0 kJ/mol. If 1 mol of liquid CCl4 has an entropy of 214 J/K, what is the entropy of 1 mol of the vapor at this temperature?

32

– In other words, 1 mol of CCl4 increases in entropy by 144 J/K when it vaporizes.

– The entropy of 1 mol of vapor equals the entropy of 1 mol of liquid (214 J/K) plus 144 J/K.

K)J/(mol 358144)J/K (214 vapor of Entropy

A Problem To Consider• The heat of vaporization, Hvap of carbon

tetrachloride, CCl4, at 25 oC is 43.0 kJ/mol. If 1 mol of liquid CCl4 has an entropy of 214 J/K, what is the entropy of 1 mol of the vapor at this temperature?

33

– Recall that the heat at constant pressure, qp , equals the enthalpy change, H.

H.= qp

– The second law for a spontaneous reaction at constant temperature and pressure becomes

TH

T

qS p (Spontaneous reaction, constant T and P)

Entropy, Enthalpy, and Spontaneity• Now you can see how

thermodynamics is applied to the question of reaction spontaneity.

34

– Rearranging this equation, we find

– Thus for a reaction to be spontaneous, H-TS must be negative.

– If H-TDS is positive, the reverse reaction is spontaneous.

– If H-TS=0, the reaction is at equilibrium

0STH (Spontaneous reaction, constant T and P)

Entropy, Enthalpy, and Spontaneity

35

• Spontaneity is favored by:– Release of energy

• Exothermicity

– Increasing disorder in a system• Ex: dissolving, evaporating

Entropy, Enthalpy, and Spontaneity

36

– When temperature is raised, however, the substance becomes more disordered as it absorbs heat.

– The entropy of a substance is determined by measuring how much heat is required to change its temperature per Kelvin degree.

Standard Entropies and the Third Law of Thermodynamics• The third law of

thermodynamics states that a substance that is perfectly crystalline at 0 K has an entropy of zero.

ΔS = 0 for crystal

37

– Standard state implies: –25 oC –1 atm pressure–1 M for dissolved substances.

Standard Entropies and the Third Law of Thermodynamics• The standard entropy of a

substance or ion , also called its absolute entropy, So, is the entropy value for the standard state of the species.

38

– Note that the elements have nonzero values, unlike standard enthalpies of formation, Hf

o , which by convention, are zero.



39

– The symbol So, rather than So, is used for standard entropies to emphasize that they originate from the third law.



40

)reactants(Sm)products(SnS ooo – Even without knowing the values for the entropies

of substances, you can sometimes predict the sign of So for a reaction.

Entropy Change for a Reaction• You can calculate the entropy

change for a reaction using a summation law, similar to the way you obtained o.

41

1. A reaction in which a molecule is broken into two or more smaller molecules.

2. A reaction in which there is an increase in the moles of gases.

3. A process in which a solid changes to liquid or gas or a liquid changes to gas.

– The entropy usually increases in the following situations:

Entropy Change for a Reaction

42

– The calculation is similar to that used to obtain Ho from standard enthalpies of formation.

)l(OH)aq(CONHNH)g(CO)g(NH2 22223


• Calculate the change in entropy, So, at 25oC for the reaction in which urea is formed from NH3 and CO2. The standard entropy of NH2CONH2 is 174 J/(mol.K).

43

)l(OH)aq(CONHNH)g(CO)g(NH2 22223

So: 2 x 193 214 174 70


• Calculate the change in entropy, So, at 25oC for the reaction in which urea is formed from NH3 and CO2. The standard entropy of NH2CONH2 is 174 J/(mol.K). )reactants(Sm)products(SnS ooo

J/K 356K/J)]2141932()70174[(So

44

Enthalpy v. Entropy for spontaneous reactions• Enthalpy Ho

– Measures relative to standard state

– Can be + or –ΔHo = ΣnHo

prod – ΣnHo

react

– kJ / mol rxn– Deals w/ entire

reaction

• Entropy So

– Measures relative to absolute zero

– Always +ΔSo = ΣnSo

prod – ΣnSo

react

– J / mol K– Deals w. a

substance

45

– State function– Calculates the maximum energy available

in the form of work.– Gives a direct criterion for spontaneity of

reaction.

Free Energy Concept

• The American physicist J. Willard Gibbs introduced the concept of free energy (sometimes called the Gibbs free energy), G, which is a thermodynamic quantity defined by the equation G=H-TS.

46

– Thus, if you can show that if G is negative at a given temperature and pressure, you can predict that the reaction will be spontaneous.

STHG

Free Energy and Spontaneity• Changes in H and S during a

reaction result in a change in free energy, G , given by the equation

47

Standard Free-Energy Change

• The standard free energy change, Go, is the free energy change that occurs when reactants and products are in their standard states.

48

– Place below each formula the values of Hfo and

So multiplied by stoichiometric coefficients.

)g(NH2)g(H3)g(N 322

So: 3 x 130.6191.5 2 x 193 J/K

Hfo: 00 2 x (-45.9) kJ


• WhWhat is the standard free energy change, Go, for the following reaction at 25oC? Use values of Hf

o and So, from reference tables.

49

– You can calculate Ho and So using their respective summation laws.

)g(NH2)g(H3)g(N 322


of

o kJ 8.91kJ ]0)9.45(2[


• What is the standard free energy change, Go, for the following reaction at 25oC? Use values of Hf


50

– You can calculate Ho and So using their respective summation laws.

)g(NH2)g(H3)g(N 322

)reactants(Sm)products(SnS ooo J/K -197J/K )]6.13035.191(1932[




51

– Now substitute into our equation for Go. Note that So is converted to kJ/K.

)g(NH2)g(H3)g(N 322

kJ/K) 0.197K)( (298kJ 91.8

ooo STHG

kJ 33.1



o and So, from Tables 6.2 and 19.1.

52

– By tabulating Gfo for substances you can calculate the

Go for a reaction by using a summation law.

)reactants(Gm)products(GnG of

of

o

Standard Free Energies of Formation• The standard free energy of

formation, Gfo, of a substance is

the free energy change that occurs when 1 mol of a substance is formed from its elements in their most stable states at 1 atm pressure and 25oC.

53

)g(OH3)g(CO2)g(O3)l(OHHC 22252

Gfo: -174.8 0 2(-394.4) 3(-228.6)kJ

– Place below each formula the values of Gfo

multiplied by stoichiometric coefficients.


• Calculate Go for the combustion of 1 mol of ethanol, C2H5OH, at 25oC. Use the standard free energies of formation.

54

)g(OH3)g(CO2)g(O3)l(OHHC 22252 Gf

o: -174.8 0 2(-394.4) 3(-228.6)kJ

– You can calculate Go using the summation law.


of

o kJ )]8.174()6.228(3)4.394(2[Go


• Calculate Go for the combustion of 1 mol of ethanol, C2H5OH, at 25oC. Use the standard free energies of formation given in Table 19.2.

55

– You can calculate Go using the summation law.


of

o kJ 8.1299Go



o: -174.8 0 2(-394.4) 3(-228.6)kJ


o: -174.8 0 2(-394.4) 3(-228.6)kJ

• Calculate Go for the combustion of 1 mol of ethanol, C2H5OH, at 25oC. Use the standard free energies of formation given in Table 19.2.

56

1. When Go is a large negative number (more negative than about –10 kJ), the reaction is spontaneous as written, and the reactants transform almost entirely to products when equilibrium is reached.

Go as a Criteria for Spontaneity• The following rules are useful in

judging the spontaneity of a reaction.

57

2. When Go is a large positive number (more positive than about +10 kJ), the reaction is nonspontaneous as written, and reactants do not give significant amounts of product at equilibrium.

• The following rules are useful in judging the spontaneity of a reaction.

Go as a Criteria for Spontaneity

58

3. When Go is a small negative or positive value (less than about 10 kJ), the reaction gives an equilibrium mixture with significant amounts of both reactants and products.

• The following rules are useful in judging the spontaneity of a reaction.

Go as a Criteria for Spontaneity

59

Go as a Criteria for SpontaneityIn Summary:• ΔGo < (–)10 kJ – spontaneous• ΔGo > (+)10 kJ – nonspontaneous• (-)10 kJ <ΔGo > (+)10 kJ –

equilibrium– @ equilibrium ΔGo is usually set equal

to zero.– ΔGo = 0– This assumption is useful for many

calculations.

60

ΔGo at equilibrium

At what temperature does the reactionSnO2+2CO 2CO2+Sn

reach equilibrium?Given: ΔHo=14.7 kJ/mol; ΔSo=31.21

J/mol KΔG=ΔH-TΔS

0=ΔH-TΔS ΔH = TΔS ΔH = T ΔS

61

ΔGo at equilibrium

ΔH = T ΔS 14.7 kJ/mol = T .03121 kJ/mol K 470.7 K = T 197.7oC

62

– As a spontaneous reaction occurs, the free energy of the system decreases and entropy is produced, but no useful work is obtained.

– Remember that in any spontaneous rxn. S increases, so the quantity subtracted from H increases.

Maximum Work

• Often reactions are not carried out in a way that does useful work.

63

Maximum Work

– In principle, if a reaction is carried out to obtain the maximum useful work, no entropy is produced. (S = 0)

G = H, when S = 0

64

– It can be shown that the maximum useful work, wmax , for a spontaneous reaction is G.

– The term free energy comes from this result.

Gmaxw

Maximum Work

• Often reactions are not carried out in a way that does useful work.

65

– As the reaction proceeds, the free energy eventually reaches its minimum value.

– At that point, G = 0, and the net reaction stops; it comes to equilibrium.

Free Energy Change During Reaction• As a system approaches

equilibrium, the instantaneous change in free energy approaches zero.

Figure 19.9 Free-energy change during a spontaneous reaction.

67

– During a nonspontaneous reaction– there is a small decrease in free energy as

the system goes to equilibrium.

Free Energy Change During Reaction• As a system approaches

equilibrium, the instantaneous change in free energy approaches zero.

Figure 19.10 Free-energy change during a nonspontaneous reaction.

69

– Here Q is the thermodynamic form of the reaction quotient.

QlnRTGG o

Relating Go to the Equilibrium Constant

• The free energy change when reactants are in non-standard states (other than 1 atm pressure or 1 M) is related to the standard free energy change, Go, by the following equation.

70

– G represents an instantaneous change in free energy at some point in the reaction approaching equilibrium.

QlnRTGG o



71

– At equilibrium, G=0 and the reaction quotient Q becomes the equilibrium constant K.

QlnRTGG o



72

– At equilibrium, G=0 and the reaction quotient Q becomes the equilibrium constant K.

KlnRTG0 o



73

– When K > 1 , the ln K is positive and Go is negative.

– When K < 1 , the ln K is negative and Go

is positive.

KlnRTGo

• This result easily rearranges to give the basic equation relating the standard free-energy change to the equilibrium constant.


74

)l(OH (aq)CONHNH )g(CO)g(NH2 22223

– Rearrange the equation Go=-RTlnK to give

RTG

Klno


• Find the value for the equilibrium constant, K, at 25oC (298 K) for the following reaction. The standard free-energy change, Go, at 25oC equals –13.6 kJ.

75

– Substituting numerical values into the equation,

49.5K 298K)J/(mol 31.8

J106.13Kln

3


• Find the value for the equilibrium constant, K, at 25oC (298 K) for the following reaction. The standard free-energy change, Go, at 25oC equals –13.6 kJ. )l(OH (aq)CONHNH )g(CO)g(NH2 22223

76

– Hence,


• Find the value for the equilibrium constant, K, at 25oC (298 K) for the following reaction. The standard free-energy change, Go, at 25oC equals –13.6 kJ. )l(OH (aq)CONHNH )g(CO)g(NH2 22223

249.5 1042.2eK

77

Ho So Go Description

– + – Spontaneous at all T

+ – + Nonspontaneous at all T

– – + or – Spontaneous at low T;Nonspontaneous at high T

+ + + or – Nonspontaneous at low T;Spontaneous at high T

Spontaneity and Temperature Change• All of the four possible choices of

signs for Ho and So give different temperature behaviors for Go.

78

– You get the value of GTo at any temperature T by

substituting values of Ho and So at 25 oC into the following equation.

oooT STHG

Calculation of Go at Various Temperatures• In this method you assume that Ho

and So are essentially constant with respect to temperature.

79

)g(CO)s(CaO)s(CaCO 23

So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ


• Find the Go for the following reaction at 25oC and 1000oC. Relate this to reaction spontaneity.


of

o

kJ 3.178kJ)]9.1206()5.3931.635[(

80


So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ



)reactants(Sm)products(SnS ooo K/J 0.159)]9.92()7.2132.38[(

81


So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ



oooT STHG

– Now you substitute Ho, So (=0.1590 kJ/K), and T (=298K) into the equation for Gf

o.

82


So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ



)K/kJ 1590.0)(K 298(kJ3.178GoT

kJ 9.130GoT So the reaction is

nonspontaneous at 25oC.

83


So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ



– Now we’ll use 1000oC (1273 K) along with our previous values for Ho and So.

)K/kJ 1590.0)(K 1273(kJ3.178GoT

84


So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ



– Now we’ll use 1000oC (1273 K) along with our previous values for Ho and So.

So the reaction is spontaneous at 1000oC.

kJ 1.24GoT

85


So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ

A Problem To Consider• Find the Go for the following reaction

at 25oC and 1000oC. Relate this to reaction spontaneity.

– To determine the minimal temperature for spontaneity, we can set Gfº=0 and solve for T.

o

o

S

HT

86


So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ



– To determine the minimal temperature for spontaneity, we can set Gfº=0 and solve for T.

)C 848( K 1121K/kJ 1590.0

kJ 3.178T o

87


So: 38.292.9 213.7 J/K

Hfo: -635.1-1206.9 -393.5 kJ



– Thus, CaCO3 should be thermally stable until its heated to approximately 848 oC.

88

Bond Energy (BE)

• Energy required to break one mole of covalent bonds and form products.– All substances are in the gaseous

state at constant temperature and pressure

– Bond Energy = Bond Enthalpy

– ΔHorxn= ΣBEreactants-ΣBEproducts, or

– ΔHorxn= ΣBEbroken-ΣBEformed

89

Estimate the enthalpy for the combustion of 1 mole of butane (C4H10).

• C4H10 + 13/2O2 4CO2 + 5H2O

• Write the structural formula. H H H HH C C C C H + 13/2 O=O 4O=C=O + 5H-O-H H H H H3 C-C and 10 C-H 13/2O=O 8C=O 10 H-OΔH =[3(347) +10(414) +13/2(498)]-[8(741) + 10(464)] =[1041 + 4140+3237]-[5928-4640] =8418 – 10,568 =-2150 kJ/mol

90

Operational Skills

• Calculating the entropy change for a phase transition

• Predicting the sign of the entropy change of a reaction

• Calculating So for a reaction• Calculating Go from Ho and So

• Calculating Go from standard free energies of formation

• Interpreting the sign of Go

• Writing the expression for a thermodynamic equilibrium constant

• Calculating K from the standard free energy change• Calculating Go and K at various temperatures

91

Animation: Work Versus Energy Flow

Return to slide 6

(Click here to open QuickTime animation)

92

Animation: Spontaneous Reaction

Return to slide 21

(Click here to open QuickTime animation)

93

Figure 19.4:Examples of a spontaneous and nonspontaneous process.

Return to slide 21

thermodynamics and equilibrium. 2 –from chapter 6: the change in enthalpy equals the heat of...

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