22.03.2020

1

Thermodynamics part 2

1

THERMODYNAMIC STANDARD STATE

The thermodynamic standard state of a

substance is its the most stable pure form

under standard conditions (pressure one

atmosphere and temperature 25°C or 298 K )

2

1

2

22.03.2020

2

THERMODYNAMIC STANDARD STATE

Examples of elements and compounds in their

standard states are:

▪ nitrogen, gaseous diatomic molecules, N2(g);

▪ mercury, a silver-coloured liquid metal, Hg(l);

calcium, a silvery white solid metal, Ca(s);

carbon, a greyish black solid called graphite,

C(graphite)

sodium chloride, a white solid cristal, NaCl(s)

methane colourless gas, CH4(g)

3

TSS - RULES

• 1. For a pure substance in the liquid or solid

phase, the standard state is the pure liquid or

solid.

• 2. For a gas, the standard state is the gas at

a pressure of one atmosphere; in a mixture

of gases, the sum of pressures is equal to one

atm..

• 3. For a substance in a solution, the

standard state refers to one-molar

concentration.

4

3

4

22.03.2020

3

Quiz

Q.1.Could you listen other elements in their

standard state (conditions):

Q.2. The values associated with the term,

"standard reference conditions or shortly standard

condition“ in a thermochemical meaning refer to:

a. temperature: 0.00 K; pressure: 1.00 atm

b. temperature: 0.00 oC; pressure: 1.00 atm

c. temperature: 273.15 K; pressure: 1.00 Pa

d. temperature: 298.15 K; pressure: 1.00 atm

e. temperature: 298.15 K; pressure: 1.00 Pa

f. temperature: 25oC; pressure: 1 atm

5

Quiz

Choose the answer/s which present TSS conditions:

Answer

Solute/ mixture Condition

a) NaOH(aq) C = 0.1 M

b) Mixture of O2 and N2 Po2 = 0.7atm; PN2 = 0.3atm

c) He and Ar Po2 = 0.8 atm; PN2 = 1.3atm

d) HClaq C = 1 M

6

5

6

22.03.2020

4

STANDARD CHANGE

In comparison and tabulation, we often refer to

thermochemical or thermodynamic changes “at

standard states” or, more simply, to a standard

change.

To indicate a change at standard conditions, we

add a superscript zero; e.g. ∆H0 .

7

At constant T and P

When we describe a process as taking place “at

constant T and P,” we mean that the initial

and final conditions are the same.

Because we are dealing with changes in state

functions, the net change is the same as the

change we would have obtained hypothetically

with T and P actually held constant.

8

7

8

22.03.2020

5

STANDARD MOLAR ENTHALPIES OF FORMATION, ∆H

It is not possible to determine the total enthalpy

content of a substance on an absolute scale.

We could describe only changes in this state

function.

9

Standard molar enthalpy of formation

The standard molar enthalpy of formation,

∆Hf0 , of a substance is the enthalpy change

for the reaction in which one mole of the

substance in a specified state is formed from its

elements in their standard states.

By convention, the ∆Hf0 value for any element

in its standard state is zero.

10

9

10

22.03.2020

6

Interpretation of Hf0

The standard molar enthalpy of formation of

ethanol, C2H5OH(l), is - 277.7 kJ/mol.

Write the thermochemical equation for the

reaction for which Hf0 rxn - 277.7 kJ/mol rxn.

11

Solution

2 C(graphite) + 2H2(g) + ½ O2(g) → C2H5OH(l)

ΔHf0 = -277.7 kJ/mol rxn

12

11

12

22.03.2020

7

Example to solve

Q.3. The standard molar enthalpy of formation of

aluminium oxide, Al2O3(s), is -1675.7 kJ/mol. Write

the thermochemical equation for the reaction for

which Hf0 rxn -1675.7 kJ/mol rxn.

13

2Al(s) + 3/2 O2(g) → Al2O3(s)

ΔHf0 = -1675.7 kJ/ mol

Quiz

Q.4. The standard molar enthalpy of formation of hydrogen iodide, HI(g), is 26.48 kJ/mol. Write the thermochemical equation for the reaction for which Hf

0

rxn 26.48 kJ/mol rxn.

14

13

14

22.03.2020

8

HESS’S LAW

In 1840, G. H. Hess (1802–1850) published his law of heat summation, which he derived on the basis of numerous thermochemical observations.

The enthalpy change for a reaction is the same whether it occurs by one step or byany series of steps.

15

INTERPRETATION

As an analogy, consider traveling fromKansas City (elevation 884 ft (269.64m) above sea level) to Denver (elevation 5280 ft, (1609.34m)).

The change in elevation is (5280 - 884) ft= 4396 ft (1339.7 m), regardless of theroute taken.

16

15

16

22.03.2020

9

Hess law examples17

Method I

NH3(g) + HCl(g) --> NH4Cl(s) H = -176.15 kJ/mol

NH4Cl(s) + H2O(l) -->NH4Cl(aq) H = +16.32 kJ/mol

H = -159.83 kJ/mol

Starting from gaseous ammonia, hydrogen chloride and water

we obtaining the same product with the use of two methods

17

18

22.03.2020

10

Method II

NH3(g) + H2O(l) --> NH3(aq) H = -35.15 kJ/mol

HCl(g) + H2O(l) --> HCl(aq) H = -72.38 kJ/mol

NH3(aq) + HCl(aq) --> NH4Cl(aq) H = -52.30 kJ/mol

H = -159.83 kJ/mol

Hess’s Law

Hess’s law lets us calculate enthalpy changes for reactions for which the changes could be measured only with difficulty, if at all.

In general terms, Hess’s Law of heat summation may be graphicallyrepresented as follows: (next slide)

20

19

20

22.03.2020

11

A schematic representation of Hess’s Law

21

EXAMPLE

Consider the following reaction.

C (graphite) + 1/2 O2(g) →CO(g)

∆H0 rxn __ ?

The enthalpy change for this reaction cannot be

measured directly.

Even though CO(g) is the predominant product of

the reaction of graphite with a limited amount of

O2(g), some CO2(g) is always produced as well.

22

21

22

22.03.2020

12

C(graphite) + O2(g) → CO2(g)

ΔH0rxn = -393.5 kJ/mol rxn (1)

CO(g) + ½ O2(g) → CO2(g)

ΔH0rxn = -283.0 kJ/mol rxn (2)

23

Revise equation (2) to give (-2) thus we

obtain the CO(g) from carbon dioxide,

when the equation is revised, the sign of

ΔH0 is changed because the reversed of

exothermic reaction is endothermic

reaction, then add to eq.(1).

Plan 24

23

24

22.03.2020

13

EXAMPLE

C(graphite) + O2(g) → CO2(g) ΔH0rxn = -393.5 kJ/mol rxn (1)

CO2(g) → CO (g) + ½ O2(g) ΔH0rxn = -(-283.0 kJ/mol rxn) (2)

C(graphite) + ½ O2(g) → CO(g) ΔHrxn0 = - 110.5 kJ/mol rxn

25

INTERPRETATION26

25

26

22.03.2020

14

Example

Calculate heat of reaction at 298K.

C2H4(g) + H2O(l) → C2H5OH(l)

use the following thermochemical equationsΔH0

C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)

-1367kJ/mol (1)

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)

- 1411kJ/mol (2)

27

Plan

Revise equation (1) to give (-1) when the equation is revised, the sign of ΔH0 is changed because the reversed of exothermic reaction is endothermic reaction, then add to eq.(2).

28

27

28

22.03.2020

15

Solution

ΔH0

2CO2(g) + 3H2O(l) → C2H5OH(l) + 3O2(g)

+1367kJ/mol (-1)

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)

- 1411kJ/mol (2)

C2H4(g) + H2O(l) → C2H5OH(l)ΔH0 = -44 kJ/mol rxn

29

What is the value for ΔH for the following reaction?

N2(g) + 5O2(s) → 5N2O5(g)

Given:2 H2(g) + O2(g) →2 H2O(l) ΔHo/kJ = -571.6

N2O5(g) + H2O(l) →2 HNO3(l) ΔHo/kJ = -73.7

N2(g) + 3 O2(g) + H2(g) →2 HNO3(l)

ΔHo/kJ = -348.2

30

29

30

22.03.2020

16

Quiz

Calculate the standard enthalpy of formation of glucose (photosynthesisprocess) according to the followingreaction:

6CO2+ 6H2O → C6H12O6 + 6 O2

The required values of ΔHf0 are presented

above:

β-D-glucose = -1268 kJ/mol

CO2(g) = -393 kJ/mol

H2O(l) = -285,5 kJ/mol

O2(g) = 0 kJ/mol

31

CHANGES IN INTERNAL ENERGY, E

The internal energy, E, of a specific amount of a substance represents all the energies contained within the substance.

It includes such forms of energy as:

kinetic energies of the molecules;

energies of attraction, repulsion among subatomic particles, atoms, ions, ormolecules;

other forms of energy.

32

31

32

22.03.2020

17

CHANGES IN INTERNAL ENERGY, E

The internal energy of a collection of molecules

is a state function.

The difference between the internal energy of

the products and the internal energy of the

reactants of a chemical reaction or physical

change, E, is given by the equation

33

CHANGES IN INTERNAL ENERGY, E

Δ E = Efinal - Einitial - Eproducts – Ereactants - q + w

34

33

34

22.03.2020

18

Sign conventions for q and w.35

Entropy, free energy, spontaneousand non-spontaneous reaction

36

35

36

22.03.2020

19

Reaction spontaneity

Inside a pile of oily rags or a stack of hay that has not

been thoroughly dried, decomposition causes heat to build

up. When heat cannot escape, the temperature can

become high enough to cause a fire.

What was the character of the process?

Answer - spontaneous

You will learn about the conditions that will produce

a spontaneous chemical reaction.

37

Spontaneous reaction

• What are two characteristics of spontaneous reactions?

• A spontaneous reaction occurs naturally and favours the formation of products at the specified conditions.

38

37

38

22.03.2020

20

Any process which occurs without outside

intervention is spontaneous.

➢When two eggs are dropped they

spontaneously break.

➢The reverse reaction (two eggs leaping into

your hand with their shells back intact) is not

spontaneous.

39

Spontaneous vs. nonspontaneous

The reverse of any spontaneous change is nonspontaneous, because if it did occur, the universe would tend toward a state of greater order. This is contrary to our experience.

40

39

40

22.03.2020

21

Spontaneous reactions

We can conclude that a spontaneous process has

a direction.

A process that is spontaneous in one

direction, is not spontaneous in the opposite

one.

The direction of a spontaneous process can

depend on temperature.

Ice turning to water is spontaneous at T > 0°C.

Water turning to ice is spontaneous at T < 0 °C

41

Spontaneous process (irreversible)

Spontaneous processes:

✓water flowing in the water flow,

✓gas combustion in gas oven,

✓melting of ice cube in glass of cold drink

during hot day,

✓ synthesis water from hydrogen and oxygen

41

42

22.03.2020

22

Irreversible (spontaneous) process

Another examples of irreversibleprocesses:

• Expansion of gas into vacuum

• When mass flows between tworegions of various concentrations

• When heat flows between two regions of various temperature

• Each chemical reaction

A reversible process is one which can go back and

forth between states along the same path.

▪When 1 mole of water is frozen at 1 atm at 0 °C to

form 1 mole of ice, q = ∆Hsol of heat is removed.

▪To reverse the process, q = ∆Hmelt must be added to

the 1 mole of ice at 0°C and 1 atm to form 1 mole of

water at 0 °C

▪ΔHmelt = ΔHsol = 6.008 kJ/mol

44

43

44

22.03.2020

23

Although a spontaneous reaction might occur rapidly, thermodynamic spontaneity is not related to speed. The fact that a process is spontaneous does not mean that it will occur at an observable rate.

It may occur rapidly, at a moderate rate, or even very slowly.

The rate at which a spontaneous reaction occurs is addressed by kinetics (see in future lectures).

45

Spontaneous processes might be slow or fast

Gas combustion as the example of fast and spontaneous, explosive reaction .

Metal corrosion. Spontaneous and slow Reaction.

45

46

22.03.2020

24

The major concern of thermodynamics is predicting whether a particular process can occur under specified conditions to give predominantly products.

47

Product favored

A change for which the collection of products is thermodynamically more stable than the collection of reactants under the given conditions is said to be product favoured, or spontaneous, under those conditions.

48

47

48

22.03.2020

25

Reactant- favored

A change for which the products arethermodynamically less stable than the reactants under the given conditions is described as reactant-favoured, or nonspontaneous, under those conditions.

49

Non spontaneous processes

Thus non-spontaneous process does not favour products forming under specific conditions.

49

50

22.03.2020

26

Examples of non-spontaneous reactions: photosynthesis (requires extra an inpunt of externalenergy).

A Non-natural Process: Steinberg's Famous New Yorker Cartoon

Source: Boundless. “Spontaneous and Nonspontaneous Processes.” Boundless Chemistry. Boundless, 21 Jul. 2015. Retrieved 16 Sep. 2015 from https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/thermodynamics-17/the-laws-of-thermodynamics-123/spontaneous-and-nonspontaneous-processes-497-3512/

Electrolysis of water

https://pl.wikipedia.org/wiki/Plik:Elektroliza_wody.svg

51

52

22.03.2020

27

Quiz

Q. Non-spontaneous changes are (choose

proper answer):

➢ rusting of an iron

➢burning a piece of paper

➢melting of ice at room temperature

➢electrolysis

➢photosynthesis

➢2H2O→2H2 +O2

➢producing the aniline from nitrobenzene

➢Solidification of water at room temperature

53

Spontaneus change

Any spontaneous change has a natural direction, like:

the rusting of a piece of iron,

the burning of a piece of paper,

or the melting of an ice at roomtemperature.

We can think of a spontaneous process as one for which products are favoured over reactants at the specified conditions.

54

53

54

22.03.2020

28

Enthalpy of exothermic reaction

Many product-favoured reactions are exothermic.

➢ The combustion (burning) reactions of hydrocarbons such as methane and octane are all exothermic and highly product-favoured (spontaneous).

The enthalpy contents of the products are lower than those of the reactants

55

Not all exothermic changes are spontaneous, however, nor all spontaneous changes areexothermic.

As an example, consider the freezing of water, which is an exothermic process (heat is released).

This process is spontaneous at temperatures below 0°C, but it certainly is not spontaneous at temperatures above 0°C.

56

55

56

22.03.2020

29

Likewise, we can find conditions at which the melting of ice, an endothermic process, is spontaneous.

Spontaneity is favoured but not required when heat is released during a chemical reaction or a physical change.

57

The dissolution of ammonium nitrate,

NH4NO3, in water is spontaneous,

although it is the endothermic reaction.

58

57

58

22.03.2020

30

The system (consisting of the water, the

solid NH4NO3, and the resulting hydrated

NH4+ and NO3

- ions) absorbs heat from

the surroundings as the endothermic

process occurs.

59

Nevertheless, the process is spontaneous

because the system becomes more

disordered as the regularly arranged ions

of crystalline ammonium nitrate become

more randomly distributed hydrated ions

in solution.

60

+

59

60

22.03.2020

31

An increase in disorder in the system

favours the spontaneity of a reaction.

In this particular case, the increase in

disorder overrides the effect of

endothermicity.

61

Factors affect the spontaneity

Two factors affect the spontaneity of any physical or chemical change:

Spontaneity is favoured when

1. heat is released during the change (exothermic, ΔH < 0)

2. the change causes an increase in disorder.

62

61

62

22.03.2020

32

Entropy

The thermodynamic state function entropy, S, is a measure of the disorder of the system.

The greater the disorder of a system, higher is its entropy.

Now lets show some examples of systemsof increasing entropy.

63

ENTROPY - SOLID-LIQUID-GAS

For a given substance, the entropy of the gas is greater than the entropy of the liquid or the solid. Similarly, the entropy of the liquid is greater than that of the solid.

64

63

64

22.03.2020

33

INCREASING ENTROPY

Entropy increases when a substance is divided into parts.

65

ENTROPY IN CHEMICAL REACTION

Entropy tends to increase in chemical reactions in which the total number of product molecules is greater than the total number of reactant molecules.

66

65

66

22.03.2020

34

INCREASING ENTROPY

Entropy tends to increase when temperature increases. As the temperature increases, the molecules move faster and faster, which increases the disorder.

67

Entropy quiz

Q. Which ones of the processes below are not

accompanied by an increase in entropy?

a. Preparing a soup

b. decomposition of SO3

2SO3→ 2SO2 + 2O2

d. Cooling the tea in cup

e. Cleaning the floor after the party

f. Forming a snowman

g. Breaking the glass

68

67

68

22.03.2020

35

Quiz

Q. Which system has the greater enthropy:

A) granulated sugar or icing sugar

B) wall of bricks or several bricks

C) desk of pedant person or desk of

messy person

D) the lecture room before or after the

lecture

69

Quiz

Based on the relationship between entropy and disorder, indicate which, if any, of the following changes represents an increase in the entropy of the system.

a. the freezing of acetic acidb. the sublimation of the moth repellent para-dichlorobenzene (C6H4Cl2)

c. the burning of gasoline

70

69

70

22.03.2020

36

In which of the following reactions will

the entropy of the system increase?

a. CH3OH(l) → CH3OH(g)

b. N2O4(g) → 2 NO2(g)

c. 2 KClO3(s) → 2 KCl(s) + 3 O2(g)

71

Quiz

ENTROPY

• What part does entropy play in chemical reactions?

• Entropy is a measure of the disorder of a system.

◦ Physical and chemical systems attain the lowest possible energy.

◦ The law of disorder states that the natural tendency for systems is to move in the direction of maximum disorder or randomness.

72

71

72

22.03.2020

37

ENTROPY

• An increase in entropy favours the spontaneous chemical reaction.

• A decrease favours the nonspontaneous reaction.

73

ENTROPY CHANGES IN CHEMICAL REACTIONS

∆So = ∑nSoproducts - ∑ nSo

reactants

74

73

74

22.03.2020

38

What is the value ΔSrxn for the dissolutionof sodium sulphate if

Na2SO4(s)→ 2 Na+(aq) + SO2-

4(aq)

ΔS0salt = 149.6 J mol-1K

ΔS0sodium = 59.0 J mol-1K

ΔS0sulfate = 20.0 J mol-1K

75

Δ𝑆 = 2 ∙ 59 + 20 - 149.6 = - 12 J/mol K

Second law of thermodynamicsintroduction

In spontaneous changes, the universe tends toward a state of greater disorder.

The Second Law of Thermodynamics is based on our experiences.

State of entropy of the entire universe, as a closed isolated system, will always increase over time. The second law also states that the changes in the entropy in the universe can never be negative.

76

75

76

22.03.2020

39

77

78

77

78

22.03.2020

40

Enthalpy and entropy

Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = −∆𝐻𝑠𝑦𝑠𝑡𝑒𝑚

𝑇

[𝐸𝑛𝑡ℎ𝑎𝑙𝑝𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 [𝐽 𝑚𝑜𝑙−1]

𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟e [𝐾]]

79

Reversible and irreversible processes

∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚=𝑄

𝑇

Reversible process

Irreversible process, spontaneous

∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚>𝑄

𝑇

79

80

22.03.2020

41

Entropychange

Process Process kind

ΔS > 0 Process happened irreversible, spontaneous

ΔS = 0

Process is at equilibrium and it is

impossible to change its direction

reversible

ΔS < 0 Process is not happened

Non-spontaneous

If protein undergoes denaturation at 80oC, the standard enthalpy of this process is equal to 500 kJ/mol.

Calculate entropy of this process. Process undergoes at constant pressure it means that Q= ΔH.

ΔS = ∆𝐻

𝑇=

500 000 Τ𝐽 𝑚𝑜𝑙

273.14+80 𝐾= 1.42 J/mol K

81

82

22.03.2020

42

QUIZ

Change of enthalpy of carbon burningin oxygen is equal to -393,5 kJ/mol. Calculate the change of entropy of surroundings for this reaction. The temperature of the reaction is 298 K.

83

Δ𝑆surr= − Δ𝐻/𝑇

Δ𝑆surr= − (−393.5 𝑘𝐽/𝑚𝑜𝑙)/(298 𝐾)

ΔS = 1.32kJ/mol K

Calculate total entropy ΔStotal of reactionwhich undergoes at 25oC

2 CuO → 2Cu(S) + O2(g)

1. Find ΔSrxn (entropy of system) using

molar entropy values

Cu, ΔS = 33.15 J/molK

CuO, ΔS = 42.63 J/molK

O2, ΔS = 205.14 J/molK

∆So = ∑nSoproducts - ∑ nSo

reactants= 186.18 J/K

83

84

22.03.2020

43

2. Find ΔHrxn 2 CuO → 2Cu(S) + O2(g)

Using molar enthalpy values

Cu, ΔH = 0 J/mol

CuO, ΔH = -157.3 kJ/mol

O2, ΔH = 0 J/mol

∆Ho = ∑nHoproducts- ∑ nHo

reactants

ΔH0 = [(2 x 0) + (1x0)] – [- (2 mol x157.3 kJ/mol) ] = 314.6 kJ

85

3. Calculate ΔSsurr using ΔH0f and T

ΔSsurr = -314.6 𝑘𝐽

298 𝐾= - 1055.7 J/K

Δ𝑆𝑡𝑜𝑡 = Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠

Stot = 186.18 J/ K + (- 1055.7 J/K)

ΔStot = -869.52 J/K

ΔS tot is negative so reaction is non-spontaneous

85

86

22.03.2020

44

Enthalpy, Entropy, and Free Energy

• What two factors determine the spontaneity of a

reaction?

• The size and direction of enthalpy changes and

entropy changes together determine whether

a reaction is spontaneous;

• that is, whether it favours products and releases

free energy.

87

Enthalpychange

Entropy Spontaneityreaction

Decreases(exothermic)

Increases (more disorder in products than in reactants)

Yes

Increases(endothermic)

Increases Only if unfavourableenthalpy change isoffset by favourableentropy change

Decreases(exothermic)

Decreases (moredisorder in reactantsthan in products)

Only if unfavourableentropy change isoffset by favourableenthalpy change

Increases(endothermic)

Decreases No

Influence of enthalpy and entropy changes on reaction spontaneity

88

87

88

22.03.2020

45

Free Energy and SpontaneousReactions

Spontaneous reactions produce substantial amounts of products at equilibrium and release free energy.

Free energy is energy that is available to do work.

89

Gibbs free-energy

• The Gibbs free-energy change (ΔG) is the maximum amount of energy that can be coupled to another process to do useful work.

Is the Gibbs free-energy change positive or negative in a spontaneous process?

• The numerical value of ΔG is negative in spontaneous processes because the system loses free energy.

90

89

90

22.03.2020

46

The following data are given for the sublimation of naphthalene

[C10H8(s)→ C10H8(g)]

at 25°C: ΔS° = 168.7 J.K-1.mol-1; ΔH° = 73.6 kJ.mol-1. Determine ΔG° for this process at this temperature

91

Plan

1. Convert oC into K

25 oC + 273 = 298 K

2. Present ΔH in J

73.6 kJ.mol-1 = 73600 J mol-1

3. Use formula

ΔG = 73600 J mol-1 - 298 K . 168.7 J.K-1.mol-1

ΔG = 73600 J mol-1 – 50272.6 J mol-1

ΔG = 23327.4 J mol-1 = 23.33 kJ mol-1

92

91

92

22.03.2020

47

94

Free Gibbs energy in questions95

94

95

22.03.2020

48

Combustion of acetylene

C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g)

Enthalpies of formation ∆Horxn = -1238 kJ

Standard molar entropies

∆Sorxn = -97.4 J/K or -0.0974 kJ/K

∆Gorxn = -1238 kJ - (298 K)(-0.0974 J/K)

= -1209 kJ

Reaction is product-favored in spite of negative

∆Sorxn.

Reaction is “enthalpy driven”

96

Calculating ∆Gorxn

From tables of thermodynamic data we find

∆Horxn = +25.7 kJ

∆Sorxn = +108.7 J/K or +0.1087 kJ/K

∆Gorxn = +25.7 kJ - (298 K)(+0.1087 kJ/K)

= -6.7 kJ

Reaction is product-favored in spite of positive

∆Horxn.

Reaction is “entropy driven”

NH4NO3(s) + heat ---> NH4NO3(aq)

97

96

97

22.03.2020

49

:

In the Haber process for the manufacture of ammonia

N2 + 3H2→ 2 NH3 At what temperature will the reaction above become

spontaneous?

If the enthalpy and entropy of this proces

ΔH = -93 kJ/mol; ΔS= -198 J/ mol K

The fact that both terms are negative means that the Gibbs

free energy equation is balanced and temperature

dependent:

ΔG = ΔH - TΔS

98

ΔG = -93000 - (T x -198)note that the enthalpy is given in kilojoules

if ΔG = 0then the system is at the limit of reaction spontaneity

When ΔG = 0 then (T x -198) = -93000

and T = 93000/198 Kelvin

therefore the reaction becomes spontaneous whenT = 469 K (196 ºC)

below this temperature the reaction is spontaneous.

99

98

99

22.03.2020

50

Standard free energy change100

∆𝐺0 = 𝑠𝑢𝑚 𝑜𝑓 ∆𝐺𝑓0 𝑜𝑓 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 -

𝑠𝑢𝑚 𝑜𝑓 ∆𝐺𝑓0 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠

101

100

101

22.03.2020

51

What is ΔG0 for the combustion of liquid ethanol to give CO2(g) and H2O(g)?

C2H5OH(l) + 3O2(g)→ 2CO2(g) + 3H2O(g)

102

ΔG0 ethanol ΔG0 CO2 ΔG0 H2O ΔG0 O2(g)

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

∆𝐺0= 2 𝑚𝑜𝑙 ∙−394.4 𝑘𝐽

𝑚𝑜𝑙+ 3 𝑚𝑜𝑙 ∙

−228.6 𝑘𝐽

𝑚𝑜𝑙

- 1 𝑚𝑜𝑙 ∙−174.8 𝑘𝐽

𝑚𝑜𝑙+ 3 𝑚𝑜𝑙 ∙

0 𝑘𝐽

𝑚𝑜𝑙

∆𝐺0 = -1299.8 kJ

∆𝐺0 = maximum work available

103

102

103

22.03.2020

52

Free Gibbs energy and spontaneity of reaction

• ΔG < 0 The reaction is spontaneous in the forward direction (and nonspontaneous in

the reverse direction).

• ΔG > 0 The reaction is nonspontaneous in the forward direction (and spontaneous in

the reverse direction).

• ΔG = 0 The system is at equilibrium.

104

105

104

105

22.03.2020

53

Free Gibbs energy and equilibrium constant

ΔG0 = -RT ln K ΔG = -RT lnQ

Calculate the Kp for the following reaction:

2SO2(g) + O2(g) ↔ 2SO3(g)

∆𝐺0 for this reaction is equal to

– 1400 kJ/mol at 25oC.

ΔG0 = -RT ln Kp

-1400000 J/mol = - 8.314 J/molK (25+273)K lnKp

lnKp = 56.5

Kp = 3 x 1024

107

106

107

22.03.2020

54

Equilibrium constant, free Gibbs energy and directionof chemical reaction

K ΔG Direction of chemicalreaction

>1,0 negative Toward products

1,0 zero At equilibrium

<1,0 positive Toward reagents

The equation, Horeaction = Ho

f (products) - Hof (reactants)

is a statement of the second law of thermodynamics. ___

True and False

Quiz109

108

109

22.03.2020

55

True/False:

ΔG > 0, the process is spontaneous

True/False:

A nonspontaneous process cannot occur with external

intervention.

110

QUIZ

1. Free energy from a reaction is the amount of

energy that is

absorbed by an entropy decrease.

equal to the enthalpy change.

wasted as heat.

avaiable to do work

111

110

111

22.03.2020

56

QUIZ

2. Free energy is always available from

reactions that are

endothermic.

nonspontaneous.

at equilibrium.

spontaneous.

112

QUIZ

3. Choose the correct words for the spaces:

Spontaneous reactions produce ________

and substantial amounts of _________ at

equilibrium.

free energy, products

no free energy, reactants

free energy, reactants

no free energy, products

113

112

113

22.03.2020

57

QUIZ

4. Which of the following involves a decrease in entropy? Natural gas burns. A liquid freezes.

Dry ice sublimes. Water evaporates.

5. Both the enthalpy and the entropy are

combined to calculate the

a. reaction time

b. free energy

c. bound energy

114

QUIZ

6. A reaction is spontaneous if

enthalpy decreases and entropy increases.

enthalpy increases and entropy increases.

enthalpy decreases and entropy decreases.

enthalpy increases and entropy decreases

115

114

115

22.03.2020

58

QUIZ

6. Choose the correct words for the spaces: Gibbs free-energy change is the _________ amount of energy that can be ___________ another process to do useful work.

maximum, coupled to

maximum, duplicated by

spontaneous, coupled to

minimum, duplicated by

116

116