1 Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved

1

Chapters 29 and 35

Thermochemistry

and

Chemical Thermodynamics

Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved.

2

Thermochemistry

• Thermochemistry is the study of the energy effects that accompany chemical reactions.

• Why do chemical reactions occur? What is the driving force of rxn?

• Answer: Stability, wants to get to lower E. For a rxn to take place spontaneously the products of reaction must be more stable (lower E) than the starting reactants. Nonspontaneous means never happen by self.

E

R

P

release E, spon

higher E, less stable, more reactive

E R

Pabsorb E, nonspon

lower E, more stable, less reactive

3

29.1 Reaction Enthalpy• In chemical reactions, heat is often transferred from

the “system or reaction” to its “surroundings,” or vice versa.

• system - the substance or mixture of substances under study in which a change occurs.

• The surroundings are everything in the vicinity of the thermodynamic system.

system or rxn

surroundings

(

+ into system

- out system

4

Heat of Reaction• Heat flow is defined as the energy that flows

into or out of a system. We follow heat flow by watching the difference in temperature between the system and its surroundings.

• Often we follow the surroundings temp (solvent) and must realize that the opposite is happening to the system. If system is absorbing heat from the surroundings than the temp of the surroundings must be decreasing.

Tsystem (+) Tsurr (-)

5

Heat of Reaction• Heat flow or heat of reaction is denoted by the

symbol q and is the amount of heat required to return a system to the given temperature at the completion of the reaction. For an endothermic rxn the sign of q is positive;

heat is absorbed by the system from the surroundings.

E

Pabsorb heat, nonspon (endo)

R

q > 0

Surroundings

+q

Tsystem

Tsurr

System

6

Heat of Reaction

q < 0

-q

System

Surroundings

For an exothermic rxn, the sign of q is negative; heat is evolved (released) by the system to the surroundings.

Tsystem

Tsurr

E

R

P

release heat, spon (exo)

7

Enthalpy and Enthalpy Change

• The heat absorbed or evolved by a reaction depends on the conditions under which it occurs. ex. pressure

• Usually, a reaction takes place in an open vessel, and therefore under the constant pressure of the atmosphere.

• heat of this type of reaction is denoted qp; this heat at constant pressure is named enthalpy and given symbol H. H is the heat flow at constant pressure.

8

– an extensive property - depends on the quantity of substance.

– Enthalpy is a state function, a property of a system

that depends only on its present state and is independent of any previous history of the system.

Enthalpy and Enthalpy Change• Enthalpy, denoted H, is an extensive property of a

substance that can be used to obtain the heat absorbed or evolved in a chemical reaction at constant pressure.

ooo

9

• The reaction enthalpy for a reaction at a given temperature and pressure

)reactants((products) HHH


)i((final) nitialHHH

10

• As we already stated the reaction enthalpy is equal to the heat of reaction at constant pressure. This represents the entire change in internal energy (U) minus any expansion “work” done by the system; therefore we can define enthalpy and internal work by the

• 1st law of thermodynamics:• In any process, the total change in energy of the system, U, is

equal to the sum of the heat absorbed, q, and the work, w, done by the system.

• U = qp + w = H + w


11

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

– These exchanges of energy are of two kinds; heat and work - must account for both.

– 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

In chemical systems, work is defined as a change in volume at a given pressure, that is:

VPw

12

negative sign is to keep sign correct in terms of system. For expansion, V, will be a positive value but expansion involves the system doing work on the surroundings and a decrease in internal energy -- negative keeps it neg. For contraction work, V, will be a negative value but contraction involves the surroundings doing work on the system and an increase in internal energy -- negative keeps it positive (- x - = +).

Giving us the 1st law of thermo is more useful form:

VPHU realize absorb heat (+)

release or evolved heat (-)HW 44

VPw

code: first

13

29.3 Thermochemical Equations

• A thermochemical equation is the chemical equation for a reaction (including phase labels {important}) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation.

kJ -91.8H );g(NH2)g(H3)g(N 322

If H has a superscript like Ho, means thermo standard conditions -- 25oC (298K) and 1 atm.

14

• The following are two important rules for manipulating thermochemical equations:

– 1.) When a thermochemical equation is multiplied by any factor, the value of H for the new equation is obtained by multiplying the H in the original equation by that same factor.

– 2.) When a chemical equation is reversed, the value of H is reversed in sign.

Thermochemical Equations

kJ 967.4 H ; )(4)(2)(4

kJ 483.7- H ; )(2)()(2o

222

o222

gOHgOgH

gOHgOgH

kJ 483.7 H ; )()(2)(2

kJ 483.7- H ; )(2)()(2o

222

o222

gOgHgOH

gOHgOgH exo

endo

15

• Hess’s law of heat summation states that for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation is the sum of the enthalpy changes for the individual steps.

• Basically, R & P in individual steps can be added like algebraic quantities in determining overall equation and enthalpy change.

29.5 Hess’s Law

16

simple example :

Given: A + D E + C H = X kJ

2A + B 2C H = Y kJ

Question: 2D B + 2E H = ?

2A + 2D 2E + 2C H = 2X kJ

2C 2A + B H = -Y kJ

2D B + 2E H = 2X – Y kJ

_______________________________________

1. Correct side?

2. Correct # moles?

17

• For example, suppose you are given the following GIVEN data:

Hess’s Law

kJ -297H );g(SO)g(O)s(S o22

kJ 198H );g(O)g(SO2)g(SO2 o223

• use these data to obtain the enthalpy change for the following reaction?

?H );g(SO2)g(O3)s(S2 o32

x2

flip

18

• If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the third.

(2)kJ) -297(H );g(SO2)g(O2)s(S2 o22

(-1)kJ) 198(H );g(SO2)g(O)g(SO2 o322

kJ -792H );g(SO2)g(O3)s(S2 o32

HW 45

kJ -297H );g(SO)g(O)s(S o22

kJ 198H );g(O)g(SO2)g(SO2 o223

x2

flip

code: ten

19

• The standard enthalpy of formation of a substance, denoted Hf

o, is the enthalpy change for the formation of one mole of a substance in its standard state from its component elements in their standard state (298K & 1 atm).

– Note that the standard enthalpy of formation for a pure element in its standard state and H+ is zero. This means elements in their standard state has Hf

o = 0: metals - solids, diatomic gases, H+ ion.

29.6 Standard Enthalpies of Formation

(molecular scale)

Ag (s) + ½ Cl2 (g) AgCl (s) Hfo AgCl

20

• Another way to determine heat of reaction is the The law of summation of heats of formation which states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants.

is the mathematical symbol meaning “the sum of”, and n is the coefficients of the substances in the chemical equation..

)reactants()products( of

of

o HnHnH

Standard Enthalpies of Formation

21

Ex. Generic Law of Summation

aA + bB cC + dD


of

o HnHnH

22

A Problem to Consider

– What is the standard reaction enthalpy , Horxn, for

this reaction?

)g(OH6)g(NO4)g(O5)g(NH4 223 molkJ /9.45 0 3.90 8.241:ofH


of

o HnHnH

23

• Using the summation law:

– Be careful of arithmetic signs as they are a likely source of mistakes.

)reactants(Hm)products(HnH of

of

o

)]/0(5)/9.45(4[

)]/8.241(6)/3.90(4[

2233

22

molOkJmolOmolNHkJmolNH

OmolHkJOmolHmolNOkJmolNOH o

kJ 906Ho HW 46

)g(OH6)g(NO4)g(O5)g(NH4 223 molkJ /9.45 0 3.90 8.241:ofH

kJkJkJ

kJkJkJkJ

9066.1836.1089

]0)6.183[()]8.1450(2.361[

code: formation

24

– 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.

35.1.2 The Second Law of Thermodynamics

• The second law of thermodynamics addresses questions about spontaneity in terms of a quantity called entropy.

25

E

R

P

release E, spon (exo)

E R

Pabsorb E, nonspon (endo)

Most soluble salts dissolve in water spontaneously; however, most soluble salts dissolve by an endothermic process.

NH4NO3 (s) NH4+ (aq) + NO3

- (aq) H = 28.1 kJ

There is an increase in molecular disorder or randomness of the system.

Solids: high order/low disorder, high energy

Liquids: middle order/low disorder, medium energyGases: low order/high disorder, low energy

26

entropy (S) - is a thermodynamic quantity that is a measure of how dispersed the energy of a system is among the different possible ways that system can contain energy, typically in J/K units.

One example of entropy is the amount of molecular disorder or randomness in the system.

S increases as disorder increases and energy decreases

gases have high disorder, low energy

solids have low disorder, high energy

We typically follow the change in entropy in the system so we treat it as a state property and measure S = Sfinal - Sinitial

+ S = increase in entropy, i.e. disorder increased; -U

-S = decrease in entropy, ie. disorder decreased ; +U

This gets us to the second law of thermo

27

Entropy and the Second Law of Thermodynamics

• The second law of thermodynamics states that the total entropy of a system and its surroundings increases for a spontaneous process.

28

The tendency of a system to increase its entropy (+S) is the second important factor in determining the spontaneity of a chemical or physical change in addition to H.

recap:

spontaneous process: (system goes to lower energy state)

favored by -H (exo)

favored by +S (ie. increase disorder)

nonspontaneous process: (system goes to higher energy state)

favored by +H (endo)

favored by -S (ie. decrease in disorder)

Do both need to be true for spon rxn? No, remember soluble salt dissolving example. The larger term will dictate overall process.

29

– As temperature is raised the substance becomes more disordered as it absorbs heat and becomes a liquid then a gas, where entropy > 0; S increases as temp increase.

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

35.4 Third Law of Thermodynamics• The third law of thermodynamics states that the

entropy of all perfect crystalline substances approaches zero as the temperature approaches absolute zero (Kelvin).

30

– Standard state implies 25 oC (298K), 1 atm pressure, and 1 M for dissolved substances.(Thermo standard state)

35.5 Standard Reaction Entropy

• 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. Similar to heats of formation, Hf

o , except on absolute not relative scale.

31

– This means that elements have nonzero values for entropy (absolute scale), unlike standard enthalpies of formation, Hf

o , which by convention, are zero (relative scale).

Standard Entropies and the Third Law of Thermodynamics

– The symbol So, rather than So, is used for standard entropies to emphasize that they originate from the third law and absolute not relative values.

32

)reactants()products( ooo SmnSS

– 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 Hf

o.

33

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

The entropy usually increases in the following situations:

Entropy Change for a Reaction

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.

AB A + B +S

A(g) B(g) + C(g) +S

A(s) B(l) or B(g) +S

B(l) C(g) +S

34

Predict S and spon/nonspon based only on entropy for the following rxns:

C2H4 (g) + Br2 (g) BrCH2CH2Br (l)

2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (g)

C6H12O6 (s) 2 C2H5OH (l) + 2 CO2 (g)

HW 47

gas to liquid; decrease in disorder; -S; nonspon based on S only

9 mols gas to 10 mols of gas; increase in disorder; +S; spon based on S only

solid to liquid/gas (decompose); increase in disorder; +S; spon based on S only

code: ben

35

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

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

A Problem To Consider• Calculate the change in entropy, So, at 25oC

for the reaction in which urea is formed from NH3 and CO2.

Gas to liquid; decrease in disorder; predict -S

36

)()()()(2 22223 lOHaqCONHNHgCOgNH

So: 193 J/mol.K 214 174 70

A Problem To Consider


J/K 356)]/214)(1()/193)(2[(

)]/70)(1()/174)(1[(

23

222

molKJmolCOmolKJmolNH

molKJOmolHmolKJCONHmolNHS o

decrease in disorder as predicted

37

– This quantity gives a direct criterion for spontaneity of reaction.

35.6 Gibbs Free Energy

• The question arises as to how do we decide if enthalpy or entropy dictates the spontaneity of a reaction. What is the relationship between H and S?

• 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 T – Kelvin scale

38

At a given temperature and pressure

G = 0, the reaction gives an equilibrium mixture with significant amounts of both reactants and products (Temp transfer point where reaction switches spon/nonspon)

G > 0 , the reaction is nonspontaneous as written, and reactants do not give significant amounts of product at equilibrium.

G < 0 , the reaction is spontaneous as written, and the reactants transform almost entirely to products when equilibrium is reached.

STHG

Free Energy and Spontaneity• Changes in H an S during a reaction result in a change in free

energy, G , given by the equation

39

H S G Description

– (exo)spon

+disorder

spon

–

spon

Spontaneous at all T

• Lets look at relationship among the signs of H, S and G and spontaneity. Note that temperature will dictate which will rule. Also realize T is in K meaning no negative temp.

enthalpy rules at low temp but entropy at very high T

STHG

+ (endo)

non

–disorder

non

+

non

Nonspontaneous at all T

–

– (exo)

Spon

–disorder

non

+ or –

Spontaneous at low T (room); H > TS; -G

Nonspontaneous at high T (1000K); H < TS

+G

+ (endo)

Non

+disorder

spon

+ or –

Nonspontaneous at low T; H > TS; +G

Spontaneous at high T; H < TS; -G

40

– The next example illustrates the calculation of the standard free energy change, Go, from Ho and So.

ooo STHG

35.7 Gibbs Energy and Equilibrium

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

41

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

So: 130.6191.5 193 J/mol K

Hfo: 00 -45.9 kJ/mol


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

predict

H, sponS, nonspon

G, spon

42


of

o kJmolkJmolNH 8.91 ]0[)]/9.45(2[ 3


kJ/K -0.197J/K -197 )]/6.130)(3(

)/5.191)(1[()]/193)(2[(

2

23

molKJmolH

molKJmolNmolKJmolNH

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

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

kJ 33.1 spon rxn as written

)(2)(3)( 322 gNHgHgN

So: 130.6191.5 193 J/mol K

Hfo: 00 -45.9 kJ/mol

43

– 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, Gf

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

44

)(3)(2)(3)( 22252 gOHgCOgOlOHHC

Gfo: -174.8 0 -394.4 -228.6 kJ/mol

A Problem To Consider• Calculate Go for the following reaction at

25oC using std. free energies of formation.

)reactants(Gm)products(GnG of

of

o

]0)/8.174)(1[(

)]/6.228)(3()/4.394)(2[(

52

22

molkJOHHmolC

molkJOmolHmolkJmolCOGo

kJ 8.1299 oG spon rxn

45

– Here Q is the thermodynamic form of the reaction quotient ([products]/[reactants] not necessarily at equil); T in kelvin; R=8.31 J/molK.

QlnRTGG o

Relating Go to the Equilibrium Constant

• The free energy change (G) when reactants are in non-standard states (meaning other than 298K, 1 atm pressure or 1 M) is related to the standard free energy change, Go, by the following equation.

46

– G represents an instantaneous change in free energy at some point in the reaction approaching equilibrium G=0.


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

KlnRTG0 o

QRTGG o ln

47

– When K > 1 (meaning equil lies to the right), the ln K is positive and Go is negative (spon).

– When K < 1 (meaning equil lies to the left), the ln K is negative and Go is positive (nonspon).

KlnRTGo

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


48

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

– Rearrange the equation Go= -RTlnK to give

RTG

Klno

A Problem To Consider• 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/mol.

49

– Substituting numerical values into the equation,

49.5K 298K)J/(mol 31.8

/106.13ln

3

molJK


2401042.2 249.5 eK

RTG

Klno

50

– 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

• We typically assume that Ho and So are essentially constant with respect to temperature.

51

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

So: 38.292.9 213.7 J/mol K

Hfo: -635.1-1206.9 -393.5 kJ/mol


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

52


of

o kJ 3.178kJ)]9.1206()5.3931.635[(


kJ/K 0.1590/ 0.159)]9.92()7.2132.38[( KJooo

T STHG – Now you substitute Ho, So (=0.1590 kJ/K), and

T (=298K) into the equation for Gfo.

)/ 1590.0)( 298(3.17825

KkJKkJGo

Co

kJ 9.13025

o

CoG

So the reaction is nonspontaneous at 25oC.


So: 38.292.9 213.7 J/mol K

Hfo: -635.1-1206.9 -393.5 kJ/mol

53


• Find the Go for the following reaction at 1000oC. – Now we’ll use 1000oC (1273 K) along with our

previous values for Ho and So because assume does not change much.

)/ 1590.0)( 1273(3.1781000

KkJKkJGo

Co

kJ 1.241000

o

CoG So the reaction is

spontaneous at 1000oC.

You see that this reaction change from nonspon to spon somewhere between 25oC to 1000oC. How can we determine at what temp this switch occurred? G=0 is equil, switch point

54

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

o

o

oo

ooo

S

HT

STH

STHG

0

)C 848( K 1121K/kJ 1590.0

kJ 3.178T o

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

– This is way you could calculate the normal boiling point of a liquid. At G=0, the liquid phase and gas phase will be at equilibrium; temperature at which switch from liquid to gaseous phase.

HW 48

nonspon < 848oC; CaCO3 stable

spon > 848oC; CaCO3 decomposes easily


l g

code: six

Documents

1 Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics Copyright (c) 2011 by Michael A. Janusa, PhD. All rights reserved