(I) 1

Ch 10 Spontaneity, Entropy and 

Free Energy 

Copyright©2000 by Houghton 

Mifflin Company. All rights reserved. 

1

The first law of thermodynamics : law of 

conservation of energy (energy can be neither 

created nor destroyed; the energy of the universe 

is constant). 

CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O (g) + energy 



2

Figure 10.1: Methane and oxygen react to form 

carbon dioxide and water 

Potential energy is converted to thermal energy, 

but the energy content of the universe (system 

+ surrounding) remains constant. 



3

The first law of thermodynamics is used mainly 

for energy bookkeeping ( 簿記 ) – that is, to 

answer the questions such as following : 

How much energy is involved in the change ? 

Does energy flow into or out of system ? 

What form does the energy finally assume ? 



4

The first law of thermodynamics provides the 

means to account for energy changes, but no 

hint as to why a particular process occurs in a 

given direction. 



5

10.1 Spontaneous processes 

A process is said to be spontaneous if it occurs 

without outside intervention. Thermodynamics 

can tell us the direction in which a process will 

occur but can say nothing about the speed 

(rate) of the process. 



6

In describing a chemical reaction : 

Chemical kinetics – focuses on the pathway 

between reactants & products. 

Thermodynamics – considers the initial & final 

states. 



7

Figure 10.2: Rate of 

reaction depends 

on the pathway from 

reactants to products 



8

In summary : 

Thermodynamics predicts whether a process will 

occur but gives no information about the time 

required, i.e., a diamond (3 D) should change 

spontaneously to graphite (2 D) at 25°C & 1 atm. 



9



10

The characteristic common to all spontaneous 

processes is an increase in a property called 

entropy (S). The change in the entropy of the 

universe for a given process is a measure of the 

driving force behind that process. 



11

Entropy is closely associated with probability. 

The key concept is that the more ways a particular 

state can be achieved, the greater is the likelihood 

(probability) that that state will occur. 



12

In other words, nature spontaneously proceeds 

toward the states that have the highest 

probabilities of existing (quantum mechanical 

aspect). I.e., the configuration with the highest 

multiplicity is the most stable configuration 

(exchange energy). 



13

The Aufbau Principle 

1. Electrons are placed in orbitals to give the lowest 

total energy to the atom. 

2. Pauli exclusion principle. 

3. Hund’s rule of maximum multiplicity 



14



15

Π c : Coulombic energy of repulsion 

Π e : exchange energy, which arises purely from 

quantum mechanical considerations. This energy 

depends on the number of possible exchanges 

between two electrons with the same energy & spin. 



16



17



18

The Coulombic energy, Π c , is positive & is nearly 

constant for each pair of electrons. The exchange 

Energy, Π e , is negative & is also nearly constant 

for each possible exchange of electrons with the 

same spin. When the orbitals are degenerate ( 簡併 ), 

both favor the unpaired configuration. 



19

Figure 10.3: Expansion of an ideal gas into an 

evacuated bulb 



20

Figure 10.4: 

Three possible 

arrangements 

(states) of four 

molecules 



21

Microstates : each configuration that gives a 

particular arrangement. 

The probability of occurrence of a particular 

arrangement (state) depends on the number of 

ways (microstates) in which that arrangement 

can be achieved. 



22

Positional probability : depends on the # of 

configurations in space (positional microstates). 

A gas expands into a vacuum to give a uniform 

distribution because the expanded state has the 

highest positional probability of the states. 

S solid < S liquid



24



25



26

10.2 The isothermal expansion and 

compression of an ideal gas 

For any isothermal process involving an ideal gas 

(the energy of an ideal gas can be changed only by 

changing its temp), 

ΔE = 0 = q + w & then q = -w 



27

Figure 10.5: A device for the isothermal 

expansion/compression of an ideal gas 

A 



28

Initially, assume the gas at 1 (P 1 , V 1 ) state, where 

P 1 is just balanced by a mass M 1 on the pan. Thus 

P 1 = force/area = M 1 g/A 

State 1 (P 1 , V 1 , n, T), where n & T remain constant 

as the expansion occurs. 



29

One-step expansion – no work 

1 (P 1 , V 1 ) → 2 (P 1 /4, 4V 1 ) - M 1 removed (no mass 

on the pan) 

No work (w 0 = 0) & heat (T constant), & this is 

called a free expansion. 



30

One-step expansion 

1 (P 1 , V 1 ) → 2 (P 1 /4, 4V 1 ) – M : M 1 → M 1 /4 

P ex = (M 1 /4)g/A = P 1 /4 

⏐Work⏐=⏐w 1 ⏐= (M 1 /4)g x h 

(w = F x d –F = M 1 g/4; d = h : change in height) 

⏐w 1 ⏐= P ex ΔV= P 1 /4(V 2 -V 1 ) = P 1 /4(4V 1 -V 1 ) 

= 3/4P 1 V 1 

w 1 = -P ex ΔV= -3/4P 1 V 1 



31

Two-step expansion 

(I) 1 (P 1 , V 1 ) → 2’ (P 1 /2, 2V 1 ) – M : M 1 → M 1 /2 

P ex =(M 1 /2)g/A = P 1 /2 

⏐w’ 2 ⏐= (P 1 /2)(V 2 -V 1 ) = (P 1 /2)(2V 1 -V 1 ) = P 1 V 1 /2 



32

(II) 2’ (P 1 /2, 2V 1 ) → 2 (P 1 /4, 4V 1 ) – M : M 1 /2 → M 1 /4 

⏐w” 2 ⏐= (P 1 /4)(4V 1 -2V 1 ) = P 1 V 1 /2 

⏐w 2 ⏐= P 1 V 1 /2 + P 1 V 1 /2 = P 1 V 1 , w 2 = -P 1 V 1 



33

⏐w 2 ⏐>⏐w 1 ⏐>⏐w 0 ⏐, 

all from state 1 (P 1 , V 1 ) to state 2 (P 1 /4, 4V 1 ). 

Work is pathway-dependent – not a state function. 



34

Figure 10.6: The PV diagram for a 

two-step expansion 



35

Six-step expansion 

1 (P 1 , V 1 ) → 2 (P 2 , V 2 ) – M = M 1 -M 1 /4 

⏐w 6 ⏐> ⏐w 2 ⏐ 



36

Figure 10.7: The PV diagram for a 

six-step expansion 



37

Infinite-step expansion 

1 (P 1 , V 1 ) → 2 (P 2 , V 2 ) – M = M n 



38

Figure 10.8: The PV diagram for the 

reversible expansion 



39

It is important to recognize that when the expansion 

of the gas is carried out in an infinite # of steps, the 

external pressure is always almost exactly equal to 

the pressure produced by the gas. 

P ex ~ P gas = P 

A process always at equilibrium is called a reversible 

process. 



40



41

The maximum work that a given amount of gas 

can perform in going from V 1 to V 2 at constant 

temp occurs in the reversible expansion. 

⏐w max ⏐=⏐w rev ⏐= nRTln(V 2 /V 1 ) 

(For the isothermal expansion of n moles of an 

ideal gas.) 



42

ΔE = 0 at constant temp, q = -w. 

As the expansion occurs, and the work (w) is 

performed. : 

w rev = -nRTln(V 2 /V 1 ) 

q rev = -w rev = nRTln(V 2 /V 1 ) 



43

The isothermal compression of an ideal gas 

One-step compression : 

1 (P 1 /4, 4V 1 ) → 2 (P 1 , V 1 ) – M = M 1 

⏐w’ 1 ⏐ = M 1 gh = P 1 ΔV = P 1 (4V 1 -V 1 ) = 3P 1 V 1 

Two-step compression : 

1 (P 1 /4, 4V 1 ) → 2 (P 1 , V 1 ) - M = 1/2M 1 → M 1 

⏐w’ 2 ⏐ = (P 1 /2)(4V 1 -2V 1 ) + (P 1 )(2V 1 –V 1 ) = 2P 1 V 1 



44

Infinite-step compression 

If we compress the gas in an infinite # of steps 

(at all time, P ex ~ P), the work required is 



45

Because P ex ~ P throughput the process, this is a 

reversible compression. Thus, in the reversible and 

isothermal compression of the gas, 

q ∞ = -w’ ∞ = -1.4P 1 V 1 (ΔE=0). 

In general terms, for a reversible compression from 

V 1 to V 2 , 

w rev = -q rev = -nRTln(V 2 /V 1 ) = -nRTln(P 1 /P 2 ) 



46

Figure 10.9: The situation before the 

one-step compression 



47



48

Summary 

Only when the expansion & compression are both 

done reversibly (by an infinite number of steps) is 

the universe the same after the cyclic process (the 

expansion and the subsequent compression of the 

gas back to its original state). That is, only for the 

reversible processes is the heat absorbed during 

expansion exactly equal to the heat released 

during compression. 



49

10.3 The definition of entropy 

Entropy is related to probability. 

S = k B lnΩ 

k B : Boltzmann’s constant (R/N A = 1.38 x 10 -23 J/K) 

Ω : # of microstates corresponding to a given state 



50

Figure 10.10: A particle in a gas that expands 

from V 1 to V 2 . 



51

For V 1 → V 2 (Ω 2 = 2Ω 1 ) 



52

For 1 mole of particles : 



53



54

P. 41/42 

(Isothermal expansion : 

q = -w) 

Probability → thermodynamics 



55

10.4 Entropy and physical changes 

How the entropy of a substance depends on its 

temp & on its physical state ? 



56

(1) Temp dependence of entropy 

for n moles of substances 



57

Probability → thermodynamics 



58

(2) Entropy changes associated with changes of state 

Since a change of state from solid to liquid at the 

substance’s melting point is a reversible process, 

we can calculate the changes in entropy for this 

process by using the equation : 

ΔS = q rev /T 

(q rev = ΔH fusion or vaporization = energy required to melt 

or vaporize 1 mol of solid / liquid at the m.p. / b.p.) 



59



60



61



62

10.5 Entropy & the second law of 

thermodynamics 

Nature always moves toward the most probable state 

(spontaneous processes will result in an increase in 

disorder). 

the second law of thermodynamics - in any 

spontaneous process there is always an increase in 

the entropy of the universe (the entropy of the 

universe is increasing) 



63

ΔS univ = ΔS sys + ΔS surr 

ΔS univ > 0 : spontaneous 

ΔS univ < 0 : non-spontaneous 



64

10.6 The effect of temp on spontaneity 

H 2 O (l) → H 2 O (g) 

For positional probability, this process has entropy 

increase (ΔS sys > 0), but ΔS surr is negative due to an 

endothermic reaction (thermal energy can be 

converted into kinetic energy associated with 

randomness). 

Which one controls the situation ? 

A spontaneous process or not ? 



65

The central idea is the entropy changes in the surroundings 

are primarily determined by heat flow (q). 

The significance of endothermicity/exothermicity as a 

driving force depends on the temperature at which 

the process occurs. That is, the magnitude of ΔS surr 

depends on the temp at which the heat is transferred 

(q rev /T). 



66

The entropy changes that occur in the surroundings 

have two important characteristics : 

1. The sign of ΔS surr depends on the direction of 

the heat flow. For an exothermic reaction, 

ΔS surr > 0. An important driving force in nature 

results from the tendency of a system to achieve 

the lowest possible energy by transferring 

energy as heat to the surroundings. 



67

2. The magnitude of ΔS surr depends on the temp. 

The transfer of a given quantity of energy as 

heat produces a much greater percentage 

change in the randomness of the surroundings 

at a low temp than it does at a high temp. Thus, 

ΔS surr depends directly on the quantity of heat 

transferred & inversely on temp (α q rev /T). 



68

Driving force provided by the energy flow (heat) = 

magnitude of entropy change of the surroundings = 

quantity of heat (J) / temp (K) 

Exothermic process : 

ΔS surr = + quantity of heat (J) / temp (K) 

Endothermic process : 

ΔS surr = - quantity of heat (J) / temp (K) 



69

Heat flow (constant P) = change in enthalpy 

= ΔH (a sign & a number) 

ΔS surr = - ΔH / T 

For a reaction that takes place under conditions 

of constant temp (K) & pressure. 



70

( 銻冶金 ) 



71



72

depends on temp 

ΔS surr = -ΔH / T 



73

10.7 Free energy 

Free energy (G) is defined as : G = H – TS 

For a process that occurs at constant temp, 

the change in free energy (ΔG) is given by 

the equation : 

ΔG = ΔH - TΔS 



74

-ΔG / T = -ΔH / T + ΔS (ΔS surr = -ΔH / T) 

-ΔG / T = ΔS surr + ΔS = ΔS univ 

ΔS univ = -ΔG / T at constant T & P 

A process (at constant T & P) is spontaneous in 

the direction in which the free energy 

decreases 

(ΔS univ = -ΔG / T > 0 : the second law of 

thermodynamics). 



75

H 2 O (s) → H 2 O (l) 

At 0°C, ΔH° = 6.03 x 10 3 J/mol 

ΔS° = ΔH°/273 = 6.03 x 10 3 /273 = 22.1 JK -1 mol -1 



76

ΔG = ΔH - TΔS 



77



78



79



80

10.8 Entropy changes in chemical reactions 

The second law of thermodynamics : a process 

will be spontaneous if the entropy of the universe 

increases when the process occurs. 

For a process at constant temp & pressure, we can 

use the change in free energy of the system to 

predict the sign of ΔS univ (= -ΔG / T) & thus the 

direction in which it is spontaneous. 



81

N 2(g) + 3H 2(g) → 2NH 3(g) 

Four reactant molecules are changed to two product 

molecules, lowering # of independent units in the 

system, thus leading to lower positional disorder. 



82

Fewer gaseous molecules mean fewer possible 

configurations . 



83

In general, when a reaction involves gaseous 

molecules, the change in positional probability is 

dominated by the relative numbers of molecules 

of gaseous reactants & products. 



84

Absolute entropies & third law of thermodynamics 

A perfect crystal (i.e., N coins have only one way 

to achieve the state of all heads) represents the 

lowest possible entropy; that is, the entropy of a 

perfect crystal at 0 K is zero - the third law of 

thermodynamics. 



85

Figure 10.11: (a) An idealized perfect 

crystal of hydrogen chloride at 0 K.; (b) > 0K. 

(a) S = 0 at 0 K; (b) S > 0 above 0 K. 



86

As the temp rises above 0 K, lattice vibrations allow 

some dipoles to change their orientations, producing 

some disorder & an increase in entropy. 



87

ΔS T 1→T2 = nCln(T 2 /T 1 ) (P. 58) 

C : heat capacity (C P or C V ), depending on the 

conditions. 

ΔS = ΔH / T (at melting or boiling point, only a 

state change occurs) 



88

Because entropy is a state function of the system, 

the entropy change for a given chemical reaction 

can be calculated by taking the difference between 

the standard entropy values of the products & 

reactants : 

ΔS° reaction = ΣS° products - ΣS° reactants 



89

It is important to note that entropy is an extensive 

property (c.f. intensive property), and it depends on 

the amount of substance present. This means that the 

# of moles of a given reactant or product must be 

taken into account. 

Extensive properties : volume, weight… 

Intensive properties : temp, density... 



90

What’s the difference 

between the entropy 

of H 2(g) & H 2 O (g) ? 



91

Figure 10.12: The H 2 O molecule can 

vibrate and rotate in several ways 

In general, the more complex 

the molecule, the higher the 

standard entropy value. 



92

10.9 Free energy and chemical reactions 

For reactions (standard free energy change, ΔG°), 

the change in free energy that occurs if the reactants 

in their standard states are converted to the products 

in their standard states. 

N 2(g) + 3H 2(g) → 2NH 3(g) 

ΔG° = -33.3 kJ 



93

(I) ΔG° = ΔH° -TΔS° 

(we cannot directly measure ΔG°, as to measure 

heat flow to determine ΔH°) 

C (S) + O 2(g) → CO 2(g) 

ΔH° = -393.5 kJ, ΔS° = 3.05 J/K 

ΔG° = ΔH° -TΔS° 

= -3.935 x 10 5 J - (298K)(3.05 J/K) 

= -3.944 x 105 J = -394.4 kJ (per mole of CO 2 ) 



94



95



96



97

Like enthalpy & entropy, free energy is a state 

function ( 狀態函數 ). Thus, we can use procedures 

for finding ΔG that are similar to those for finding 

ΔH using Hess’s law (II). 



98



99



100

The above example shows kinetic rather than 

thermodynamic control of a reaction. Thermodynamically, 

diamond should change to graphite, 

but this process is too slow to be observed. 

Diamond – kinetically stable; 

Graphite – thermodynamically stable. 

(1000°C & 10 5 atm) 



101

A third method for calculating the free energy 

change for a reaction uses standard free energies 

of formation (III). 

The standard free energy of formation (ΔG f °) of a 

substance is defined as the change in free energy 

that accompanies the formation of 1 mole of that 

substance from its constituent elements with all 

reactants & products in their standard states. 



102

6C (s) + 6H 2(g) +3O 2(g) → C 6 H 12 O 6(s) 

ΔG° = ΣΔG f ° (products) - ΣΔG f ° (reactants) 

The standard free energy of formation of an 

element in its standard state is zero. 



103



104



105

10.10 The dependence of free energy on 

pressure 

For an ideal gas, enthalpy is not pressure-dependent. 

However, entropy does depend on pressure because 

of its dependence on volume. 

In summary, at a given temp for 1 mol of ideal gas 

S large volume (low pressure) > S small volume (high pressure) 



106

G = G° + RTln(P) 

G° : free energy of the gas at 1 atm 

G : free energy of the gas at P atm 

R : gas constant 

T : Kelvin temp 



107



108

Q (reaction quotient, 商 ) 

ΔG : free energy change for the reaction at the 

specified pressure of reactants & products. 



109



110



111

Note that ΔG is significantly more negative than 

ΔG°, implying the reaction is more spontaneous 

at reactant pressure greater than 1 atm (follows 

Le Châtelier’s principle). 



112

The meaning of ΔG for a chemical reaction 

The system can achieve the lowest possible free 

energy by going to equilibrium, not by going to 

completion. That is to say, the system will 

spontaneously go to the equilibrium position, the 

lowest possible free energy, but not proceed to 

pure products or remain to pure reactants. 



113

Figure 10.13: Schematic representation of 

balls rolling down two types of hills. 

Phase change : ice to water at 25°C. 

Equilibrium position 



114

10.11 Free energy and equilibrium 

The equilibrium point (equilibrium position : 

forward & reverse reaction rates are equal) occurs 

at the lowest value of free energy available to the 

reaction system. 



115

To understand the relationship between free energy 

& equilibrium, let’s consider the following simple 

hypothetical reaction : 

A (g) ⇔ B (g) (A : 1.0 mol & 2 atm) 



116

Figure 10.14: (a) The 

initial free energies of 

A and B. (b) As A (g) 

changes to B (g) , the 

free energy of A 

decreases & that of B 

increases. 

G = G° + RTln(P) 



117

Free energy of A = G A = n A (G A ° + RTln(P A )) 

Free energy of B = G B = n B (G B ° + RTln(P B )) 

Total free energy of system = G = G A + G B 

The system has reached equilibrium [10.14(c)], & 

the system has also reached minimum free energy 

(G A = G B ). 



118

Thus, there is no longer any driving force to 

change A to B or B to A, so the system remains 

at this position (the pressures of A & B remain 

constant). 



119

Figure 10.15: The change in free energy to 

reach equilibrium 

1.0 mol A (g) at 2 atm 

(reactant only) 

1.0 mol A (g) + B (g) at 2 atm 

1.0 mol B (g) at 2 atm (product only) 



120

A : (0.25)(2 atm) = 0.50 atm 

B : (0.75)(2 atm) = 1.5 atm 

K = P Be /P Ae = 1.5 atm/0.50 atm = 3.0 

In summary 

At equilibrium, ΔG = G products -G reactants = 0 

ΔG = ΔG° + RTln(Q) 

ΔG = 0 = ΔG° + RTln(K) 

ΔG° = -RTln(K) 



121

ΔG° = -RTln(K) 



122



123



124

K=1 

K>1 



125



126



127



128

The temp dependence of K 

ΔG° = -RTln(K) = ΔH° -TΔS° 

ln(K) = -ΔH°/RT + ΔS°/R = -ΔH°/R(1/T) + ΔS°/R 



129

Figure 10.16: Experimental data showing 

the dependence of K on T 



130

Van’t Hoff equation 

(ΔH° & ΔS ° are constants over the temp range) 



131



132

10.12 Free energy and work 

Qualitatively 

The (sign) change in free energy tell us whether 

a given reaction process is spontaneous. The 

goal : to find an efficient & economical reaction 

pathway from a point of view of thermodynamics 

(spontaneous or not) is useful. 



133

Quantitatively 

The change in free energy is important quantitatively 

because it can tell us how much work can 

be done through a given process. The maximum 

possible useful work obtained from a process at 

constant temp & pressure is equal to the change 

in free energy : 

w useful 

max 

= ΔG 



134

Under certain conditions ΔG for a spontaneous 

process represents the energy that is free to do 

useful work. On the other hand, for a process that 

is not spontaneous, the value of ΔG tells us the 

minimum amount of work that must be expended 

to make the process occur. 



135

To prove the relationship between ΔG & w useful max : 

w = w useful + w useless (PV work) = w useful + w PV 

ΔE = q P + w = q P + w useful + w PV 

= q P + w useful -PΔV (at constant P & T) 

(H = E + PV) 

ΔE 

ΔH = ΔE + PΔV = q P + w useful -PΔV + PΔV 

= q P + w useful 



136

(G = H – TS) 

ΔH 

ΔG = ΔH - TΔS = q P + w useful -TΔS 

For the reversible pathway : 

w useful = w 

max 

useful & q P = q 

rev 

P 

Thus for the reversible pathway 

ΔG = q 

rev 

P + w 

max 

useful -TΔS 



137

ΔS = q P 

rev 

/T & then q P 

rev 

=TΔS (P. 56) 

So that 

ΔG = TΔS + w useful 

max 

-TΔS 

or ΔG = w useful 

max 



138

(I) If a process is carried out so that w useful = 0, 

then the expression 

ΔG = q P + w useful -TΔS 

ΔG = q P –TΔS (ΔG = ΔH - TΔS) 

ΔH – TΔS = q P –TΔS 

q P = ΔH (at constant pressure & when no useful 

work is done) 



139

(II) If a process is carried out so that w useful is 

maximum (the hypothetical reversible pathway 

where ΔG = w useful ), then the expression 

q P = TΔS (ΔG = q P + w useful –TΔS) 



140

Thus q P , which is pathway-dependent, varies 

between ΔH (when w useful = 0) & TΔS (w useful = 

w useful 

max 

). 

TΔS represents the minimum heat flow that must 

accompany the process under consideration. That 

is, TΔS represents the minimum energy that must 

be “wasted” through heat flow as the process 

occurs. 



141

In summary - at constant T & P 

(I) q P = ΔH if w useful = 0 

(II) q P = TΔS if w useful = w useful 

max 



142

10.14 Adiabatic processes 

Adiabatic process (v.s. isothermal process) : a 

process in which no energy as heat flows into or 

out of the system. 

For an adiabatic process 

q = 0 & ΔE = q + w = w 



143

Figure 10.18 : An ideal gas 

confined in an isolated 

container with a movable 

piston. No heat flow with 

the surroundings can occur. 



144

In an adiabatic expansion (v.s. in an isothermal 

process) the temperature of the gas decreases (the 

average kinetic energy of sample decreases) to 

furnish the energy to do the work. 

E = nC v Τ (ideal gas, whose energy is dependent 

on temp) 

ΔE = w = -P ex ΔV= nC v ΔT 



145

For an infinitesimal adiabatic change 

dE = -P ex dV = nC v dT 

For a reversible process (P ex ~ P gas ) 

P ex = P gas = nRT/V 

Thus, for a reversible, adiabatic expansioncompression 

dE = nC v dT = -P ex dV = -P gas dV = -(nRT/V)dV 



146

-(nRT/V)dV = nC v dT 

rearrange to (Cv/T)dT = (-R/V)dV 



147



148

Ideal gas law 



149

°C 



150



151



152

For a reversible isothermal expansion at 298 K from 

P 1 = 10.0 atm & V 1 = 12.2 L to P 2 = 1.0 atm, the final 

volume is 



153

Figure 10.19 

Comparison of the 

adiabatic & 

isothermal 

expansion for ideal 

samples in which n 

= 5 & P 1 = 10.0 atm 

& P 1 = 12.2 L. 



154

For a reversible, isothermal expansion : 

P 1 V 1 = P 2 V 2 

or PV = constant 

For a reversible, adiabatic expansion : 

P 1 V 1γ =P 2 V 2γ 

or PV γ = constant 



155

(I) 1

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