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Ch 10 Spontaneity, Entropy and<br />

Free Energy<br />

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1


The first law of thermodynamics : law of<br />

conservation of energy (energy can be neither<br />

created nor destroyed; the energy of the universe<br />

is constant).<br />

CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O (g) + energy<br />

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2


Figure 10.1: Methane and oxygen react to form<br />

carbon dioxide and water<br />

Potential energy is converted to thermal energy,<br />

but the energy content of the universe (system<br />

+ surrounding) remains constant.<br />

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3


The first law of thermodynamics is used mainly<br />

for energy bookkeeping ( 簿 記 ) – that is, to<br />

answer the questions such as following :<br />

How much energy is involved in the change ?<br />

Does energy flow into or out of system ?<br />

What form does the energy finally assume ?<br />

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4


The first law of thermodynamics provides the<br />

means to account for energy changes, but no<br />

hint as to why a particular process occurs in a<br />

given direction.<br />

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10.1 Spontaneous processes<br />

A process is said to be spontaneous if it occurs<br />

without outside intervention. Thermodynamics<br />

can tell us the direction in which a process will<br />

occur but can say nothing about the speed<br />

(rate) of the process.<br />

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6


In describing a chemical reaction :<br />

Chemical kinetics – focuses on the pathway<br />

between reactants & products.<br />

Thermodynamics – considers the initial & final<br />

states.<br />

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Figure 10.2: Rate of<br />

reaction depends<br />

on the pathway from<br />

reactants to products<br />

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In summary :<br />

Thermodynamics predicts whether a process will<br />

occur but gives no information about the time<br />

required, i.e., a diamond (3 D) should change<br />

spontaneously to graphite (2 D) at 25°C & 1 atm.<br />

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10


The characteristic common to all spontaneous<br />

processes is an increase in a property called<br />

entropy (S). The change in the entropy of the<br />

universe for a given process is a measure of the<br />

driving force behind that process.<br />

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11


Entropy is closely associated with probability.<br />

The key concept is that the more ways a particular<br />

state can be achieved, the greater is the likelihood<br />

(probability) that that state will occur.<br />

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12


In other words, nature spontaneously proceeds<br />

toward the states that have the highest<br />

probabilities of existing (quantum mechanical<br />

aspect). I.e., the configuration with the highest<br />

multiplicity is the most stable configuration<br />

(exchange energy).<br />

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13


The Aufbau Principle<br />

1. Electrons are placed in orbitals to give the lowest<br />

total energy to the atom.<br />

2. Pauli exclusion principle.<br />

3. Hund’s rule of maximum multiplicity<br />

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15


Π c : Coulombic energy of repulsion<br />

Π e : exchange energy, which arises purely from<br />

quantum mechanical considerations. This energy<br />

depends on the number of possible exchanges<br />

between two electrons with the same energy & spin.<br />

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18


The Coulombic energy, Π c , is positive & is nearly<br />

constant for each pair of electrons. The exchange<br />

Energy, Π e , is negative & is also nearly constant<br />

for each possible exchange of electrons with the<br />

same spin. When the orbitals are degenerate ( 簡 併 ),<br />

both favor the unpaired configuration.<br />

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19


Figure 10.3: Expansion of an ideal gas into an<br />

evacuated bulb<br />

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20


Figure 10.4:<br />

Three possible<br />

arrangements<br />

(states) of four<br />

molecules<br />

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21


Microstates : each configuration that gives a<br />

particular arrangement.<br />

The probability of occurrence of a particular<br />

arrangement (state) depends on the number of<br />

ways (microstates) in which that arrangement<br />

can be achieved.<br />

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22


Positional probability : depends on the # of<br />

configurations in space (positional microstates).<br />

A gas expands into a vacuum to give a uniform<br />

distribution because the expanded state has the<br />

highest positional probability of the states.<br />

S solid < S liquid


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25


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10.2 The isothermal expansion and<br />

compression of an ideal gas<br />

For any isothermal process involving an ideal gas<br />

(the energy of an ideal gas can be changed only by<br />

changing its temp),<br />

ΔE = 0 = q + w & then q = -w<br />

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Figure 10.5: A device for the isothermal<br />

expansion/compression of an ideal gas<br />

A<br />

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Initially, assume the gas at 1 (P 1 , V 1 ) state, where<br />

P 1 is just balanced by a mass M 1 on the pan. Thus<br />

P 1 = force/area = M 1 g/A<br />

State 1 (P 1 , V 1 , n, T), where n & T remain constant<br />

as the expansion occurs.<br />

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29


One-step expansion – no work<br />

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

on the pan)<br />

No work (w 0 = 0) & heat (T constant), & this is<br />

called a free expansion.<br />

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30


One-step expansion<br />

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

P ex = (M 1 /4)g/A = P 1 /4<br />

⏐Work⏐=⏐w 1 ⏐= (M 1 /4)g x h<br />

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

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

= 3/4P 1 V 1<br />

w 1 = -P ex ΔV= -3/4P 1 V 1<br />

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Two-step expansion<br />

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

P ex =(M 1 /2)g/A = P 1 /2<br />

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

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

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

⏐w 2 ⏐= P 1 V 1 /2 + P 1 V 1 /2 = P 1 V 1 , w 2 = -P 1 V 1<br />

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⏐w 2 ⏐>⏐w 1 ⏐>⏐w 0 ⏐,<br />

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

Work is pathway-dependent – not a state function.<br />

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34


Figure 10.6: The PV diagram for a<br />

two-step expansion<br />

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Six-step expansion<br />

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

⏐w 6 ⏐> ⏐w 2 ⏐<br />

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Figure 10.7: The PV diagram for a<br />

six-step expansion<br />

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Infinite-step expansion<br />

1 (P 1 , V 1 ) → 2 (P 2 , V 2 ) – M = M n<br />

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Figure 10.8: The PV diagram for the<br />

reversible expansion<br />

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39


It is important to recognize that when the expansion<br />

of the gas is carried out in an infinite # of steps, the<br />

external pressure is always almost exactly equal to<br />

the pressure produced by the gas.<br />

P ex ~ P gas = P<br />

A process always at equilibrium is called a reversible<br />

process.<br />

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40


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41


The maximum work that a given amount of gas<br />

can perform in going from V 1 to V 2 at constant<br />

temp occurs in the reversible expansion.<br />

⏐w max ⏐=⏐w rev ⏐= nRTln(V 2 /V 1 )<br />

(For the isothermal expansion of n moles of an<br />

ideal gas.)<br />

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42


ΔE = 0 at constant temp, q = -w.<br />

As the expansion occurs, and the work (w) is<br />

performed. :<br />

w rev = -nRTln(V 2 /V 1 )<br />

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

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43


The isothermal compression of an ideal gas<br />

One-step compression :<br />

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

⏐w’ 1 ⏐ = M 1 gh = P 1 ΔV = P 1 (4V 1 -V 1 ) = 3P 1 V 1<br />

Two-step compression :<br />

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

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

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44


Infinite-step compression<br />

If we compress the gas in an infinite # of steps<br />

(at all time, P ex ~ P), the work required is<br />

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45


Because P ex ~ P throughput the process, this is a<br />

reversible compression. Thus, in the reversible and<br />

isothermal compression of the gas,<br />

q ∞ = -w’ ∞ = -1.4P 1 V 1 (ΔE=0).<br />

In general terms, for a reversible compression from<br />

V 1 to V 2 ,<br />

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

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Figure 10.9: The situation before the<br />

one-step compression<br />

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Summary<br />

Only when the expansion & compression are both<br />

done reversibly (by an infinite number of steps) is<br />

the universe the same after the cyclic process (the<br />

expansion and the subsequent compression of the<br />

gas back to its original state). That is, only for the<br />

reversible processes is the heat absorbed during<br />

expansion exactly equal to the heat released<br />

during compression.<br />

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10.3 The definition of entropy<br />

Entropy is related to probability.<br />

S = k B lnΩ<br />

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

Ω : # of microstates corresponding to a given state<br />

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50


Figure 10.10: A particle in a gas that expands<br />

from V 1 to V 2 .<br />

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51


For V 1 → V 2 (Ω 2 = 2Ω 1 )<br />

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For 1 mole of particles :<br />

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P. 41/42<br />

(Isothermal expansion :<br />

q = -w)<br />

Probability → thermodynamics<br />

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55


10.4 Entropy and physical changes<br />

How the entropy of a substance depends on its<br />

temp & on its physical state ?<br />

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56


(1) Temp dependence of entropy<br />

for n moles of substances<br />

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Probability → thermodynamics<br />

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58


(2) Entropy changes associated with changes of state<br />

Since a change of state from solid to liquid at the<br />

substance’s melting point is a reversible process,<br />

we can calculate the changes in entropy for this<br />

process by using the equation :<br />

ΔS = q rev /T<br />

(q rev = ΔH fusion or vaporization = energy required to melt<br />

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

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10.5 Entropy & the second law of<br />

thermodynamics<br />

Nature always moves toward the most probable state<br />

(spontaneous processes will result in an increase in<br />

disorder).<br />

the second law of thermodynamics - in any<br />

spontaneous process there is always an increase in<br />

the entropy of the universe (the entropy of the<br />

universe is increasing)<br />

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63


ΔS univ = ΔS sys + ΔS surr<br />

ΔS univ > 0 : spontaneous<br />

ΔS univ < 0 : non-spontaneous<br />

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64


10.6 The effect of temp on spontaneity<br />

H 2 O (l) → H 2 O (g)<br />

For positional probability, this process has entropy<br />

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

endothermic reaction (thermal energy can be<br />

converted into kinetic energy associated with<br />

randomness).<br />

Which one controls the situation ?<br />

A spontaneous process or not ?<br />

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The central idea is the entropy changes in the surroundings<br />

are primarily determined by heat flow (q).<br />

The significance of endothermicity/exothermicity as a<br />

driving force depends on the temperature at which<br />

the process occurs. That is, the magnitude of ΔS surr<br />

depends on the temp at which the heat is transferred<br />

(q rev /T).<br />

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66


The entropy changes that occur in the surroundings<br />

have two important characteristics :<br />

1. The sign of ΔS surr depends on the direction of<br />

the heat flow. For an exothermic reaction,<br />

ΔS surr > 0. An important driving force in nature<br />

results from the tendency of a system to achieve<br />

the lowest possible energy by transferring<br />

energy as heat to the surroundings.<br />

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2. The magnitude of ΔS surr depends on the temp.<br />

The transfer of a given quantity of energy as<br />

heat produces a much greater percentage<br />

change in the randomness of the surroundings<br />

at a low temp than it does at a high temp. Thus,<br />

ΔS surr depends directly on the quantity of heat<br />

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

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Driving force provided by the energy flow (heat) =<br />

magnitude of entropy change of the surroundings =<br />

quantity of heat (J) / temp (K)<br />

Exothermic process :<br />

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

Endothermic process :<br />

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

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Heat flow (constant P) = change in enthalpy<br />

= ΔH (a sign & a number)<br />

ΔS surr = - ΔH / T<br />

For a reaction that takes place under conditions<br />

of constant temp (K) & pressure.<br />

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( 銻 冶 金 )<br />

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depends on temp<br />

ΔS surr = -ΔH / T<br />

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10.7 Free energy<br />

Free energy (G) is defined as : G = H – TS<br />

For a process that occurs at constant temp,<br />

the change in free energy (ΔG) is given by<br />

the equation :<br />

ΔG = ΔH - TΔS<br />

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-ΔG / T = -ΔH / T + ΔS (ΔS surr = -ΔH / T)<br />

-ΔG / T = ΔS surr + ΔS = ΔS univ<br />

ΔS univ = -ΔG / T at constant T & P<br />

A process (at constant T & P) is spontaneous in<br />

the direction in which the free energy<br />

decreases<br />

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

thermodynamics).<br />

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H 2 O (s) → H 2 O (l)<br />

At 0°C, ΔH° = 6.03 x 10 3 J/mol<br />

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

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ΔG = ΔH - TΔS<br />

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10.8 Entropy changes in chemical reactions<br />

The second law of thermodynamics : a process<br />

will be spontaneous if the entropy of the universe<br />

increases when the process occurs.<br />

For a process at constant temp & pressure, we can<br />

use the change in free energy of the system to<br />

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

direction in which it is spontaneous.<br />

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N 2(g) + 3H 2(g) → 2NH 3(g)<br />

Four reactant molecules are changed to two product<br />

molecules, lowering # of independent units in the<br />

system, thus leading to lower positional disorder.<br />

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Fewer gaseous molecules mean fewer possible<br />

configurations .<br />

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In general, when a reaction involves gaseous<br />

molecules, the change in positional probability is<br />

dominated by the relative numbers of molecules<br />

of gaseous reactants & products.<br />

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Absolute entropies & third law of thermodynamics<br />

A perfect crystal (i.e., N coins have only one way<br />

to achieve the state of all heads) represents the<br />

lowest possible entropy; that is, the entropy of a<br />

perfect crystal at 0 K is zero - the third law of<br />

thermodynamics.<br />

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Figure 10.11: (a) An idealized perfect<br />

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

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

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As the temp rises above 0 K, lattice vibrations allow<br />

some dipoles to change their orientations, producing<br />

some disorder & an increase in entropy.<br />

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ΔS T 1→T2 = nCln(T 2 /T 1 ) (P. 58)<br />

C : heat capacity (C P or C V ), depending on the<br />

conditions.<br />

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

state change occurs)<br />

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Because entropy is a state function of the system,<br />

the entropy change for a given chemical reaction<br />

can be calculated by taking the difference between<br />

the standard entropy values of the products &<br />

reactants :<br />

ΔS° reaction = ΣS° products - ΣS° reactants<br />

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It is important to note that entropy is an extensive<br />

property (c.f. intensive property), and it depends on<br />

the amount of substance present. This means that the<br />

# of moles of a given reactant or product must be<br />

taken into account.<br />

Extensive properties : volume, weight…<br />

Intensive properties : temp, density...<br />

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What’s the difference<br />

between the entropy<br />

of H 2(g) & H 2 O (g) ?<br />

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Figure 10.12: The H 2 O molecule can<br />

vibrate and rotate in several ways<br />

In general, the more complex<br />

the molecule, the higher the<br />

standard entropy value.<br />

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10.9 Free energy and chemical reactions<br />

For reactions (standard free energy change, ΔG°),<br />

the change in free energy that occurs if the reactants<br />

in their standard states are converted to the products<br />

in their standard states.<br />

N 2(g) + 3H 2(g) → 2NH 3(g)<br />

ΔG° = -33.3 kJ<br />

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(I) ΔG° = ΔH° -TΔS°<br />

(we cannot directly measure ΔG°, as to measure<br />

heat flow to determine ΔH°)<br />

C (S) + O 2(g) → CO 2(g)<br />

ΔH° = -393.5 kJ, ΔS° = 3.05 J/K<br />

ΔG° = ΔH° -TΔS°<br />

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

= -3.944 x 105 J = -394.4 kJ (per mole of CO 2 )<br />

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Like enthalpy & entropy, free energy is a state<br />

function ( 狀 態 函 數 ). Thus, we can use procedures<br />

for finding ΔG that are similar to those for finding<br />

ΔH using Hess’s law (II).<br />

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100


The above example shows kinetic rather than<br />

thermodynamic control of a reaction. Thermodynamically,<br />

diamond should change to graphite,<br />

but this process is too slow to be observed.<br />

Diamond – kinetically stable;<br />

Graphite – thermodynamically stable.<br />

(1000°C & 10 5 atm)<br />

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A third method for calculating the free energy<br />

change for a reaction uses standard free energies<br />

of formation (III).<br />

The standard free energy of formation (ΔG f °) of a<br />

substance is defined as the change in free energy<br />

that accompanies the formation of 1 mole of that<br />

substance from its constituent elements with all<br />

reactants & products in their standard states.<br />

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6C (s) + 6H 2(g) +3O 2(g) → C 6 H 12 O 6(s)<br />

ΔG° = ΣΔG f ° (products) - ΣΔG f ° (reactants)<br />

The standard free energy of formation of an<br />

element in its standard state is zero.<br />

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10.10 The dependence of free energy on<br />

pressure<br />

For an ideal gas, enthalpy is not pressure-dependent.<br />

However, entropy does depend on pressure because<br />

of its dependence on volume.<br />

In summary, at a given temp for 1 mol of ideal gas<br />

S large volume (low pressure) > S small volume (high pressure)<br />

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G = G° + RTln(P)<br />

G° : free energy of the gas at 1 atm<br />

G : free energy of the gas at P atm<br />

R : gas constant<br />

T : Kelvin temp<br />

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Q (reaction quotient, 商 )<br />

ΔG : free energy change for the reaction at the<br />

specified pressure of reactants & products.<br />

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Note that ΔG is significantly more negative than<br />

ΔG°, implying the reaction is more spontaneous<br />

at reactant pressure greater than 1 atm (follows<br />

Le Châtelier’s principle).<br />

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The meaning of ΔG for a chemical reaction<br />

The system can achieve the lowest possible free<br />

energy by going to equilibrium, not by going to<br />

completion. That is to say, the system will<br />

spontaneously go to the equilibrium position, the<br />

lowest possible free energy, but not proceed to<br />

pure products or remain to pure reactants.<br />

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Figure 10.13: Schematic representation of<br />

balls rolling down two types of hills.<br />

Phase change : ice to water at 25°C.<br />

Equilibrium position<br />

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10.11 Free energy and equilibrium<br />

The equilibrium point (equilibrium position :<br />

forward & reverse reaction rates are equal) occurs<br />

at the lowest value of free energy available to the<br />

reaction system.<br />

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To understand the relationship between free energy<br />

& equilibrium, let’s consider the following simple<br />

hypothetical reaction :<br />

A (g) ⇔ B (g) (A : 1.0 mol & 2 atm)<br />

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Figure 10.14: (a) The<br />

initial free energies of<br />

A and B. (b) As A (g)<br />

changes to B (g) , the<br />

free energy of A<br />

decreases & that of B<br />

increases.<br />

G = G° + RTln(P)<br />

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Free energy of A = G A = n A (G A ° + RTln(P A ))<br />

Free energy of B = G B = n B (G B ° + RTln(P B ))<br />

Total free energy of system = G = G A + G B<br />

The system has reached equilibrium [10.14(c)], &<br />

the system has also reached minimum free energy<br />

(G A = G B ).<br />

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Thus, there is no longer any driving force to<br />

change A to B or B to A, so the system remains<br />

at this position (the pressures of A & B remain<br />

constant).<br />

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Figure 10.15: The change in free energy to<br />

reach equilibrium<br />

1.0 mol A (g) at 2 atm<br />

(reactant only)<br />

1.0 mol A (g) + B (g) at 2 atm<br />

1.0 mol B (g) at 2 atm (product only)<br />

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A : (0.25)(2 atm) = 0.50 atm<br />

B : (0.75)(2 atm) = 1.5 atm<br />

K = P Be /P Ae = 1.5 atm/0.50 atm = 3.0<br />

In summary<br />

At equilibrium, ΔG = G products -G reactants = 0<br />

ΔG = ΔG° + RTln(Q)<br />

ΔG = 0 = ΔG° + RTln(K)<br />

ΔG° = -RTln(K)<br />

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ΔG° = -RTln(K)<br />

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K=1<br />

K>1<br />

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The temp dependence of K<br />

ΔG° = -RTln(K) = ΔH° -TΔS°<br />

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

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Figure 10.16: Experimental data showing<br />

the dependence of K on T<br />

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Van’t Hoff equation<br />

(ΔH° & ΔS ° are constants over the temp range)<br />

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10.12 Free energy and work<br />

Qualitatively<br />

The (sign) change in free energy tell us whether<br />

a given reaction process is spontaneous. The<br />

goal : to find an efficient & economical reaction<br />

pathway from a point of view of thermodynamics<br />

(spontaneous or not) is useful.<br />

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Quantitatively<br />

The change in free energy is important quantitatively<br />

because it can tell us how much work can<br />

be done through a given process. The maximum<br />

possible useful work obtained from a process at<br />

constant temp & pressure is equal to the change<br />

in free energy :<br />

w useful<br />

max<br />

= ΔG<br />

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Under certain conditions ΔG for a spontaneous<br />

process represents the energy that is free to do<br />

useful work. On the other hand, for a process that<br />

is not spontaneous, the value of ΔG tells us the<br />

minimum amount of work that must be expended<br />

to make the process occur.<br />

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To prove the relationship between ΔG & w useful max :<br />

w = w useful + w useless (PV work) = w useful + w PV<br />

ΔE = q P + w = q P + w useful + w PV<br />

= q P + w useful -PΔV (at constant P & T)<br />

(H = E + PV)<br />

ΔE<br />

ΔH = ΔE + PΔV = q P + w useful -PΔV + PΔV<br />

= q P + w useful<br />

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(G = H – TS)<br />

ΔH<br />

ΔG = ΔH - TΔS = q P + w useful -TΔS<br />

For the reversible pathway :<br />

w useful = w<br />

max<br />

useful & q P = q<br />

rev<br />

P<br />

Thus for the reversible pathway<br />

ΔG = q<br />

rev<br />

P + w<br />

max<br />

useful -TΔS<br />

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137


ΔS = q P<br />

rev<br />

/T & then q P<br />

rev<br />

=TΔS (P. 56)<br />

So that<br />

ΔG = TΔS + w useful<br />

max<br />

-TΔS<br />

or ΔG = w useful<br />

max<br />

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138


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

then the expression<br />

ΔG = q P + w useful -TΔS<br />

ΔG = q P –TΔS (ΔG = ΔH - TΔS)<br />

ΔH – TΔS = q P –TΔS<br />

q P = ΔH (at constant pressure & when no useful<br />

work is done)<br />

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139


(II) If a process is carried out so that w useful is<br />

maximum (the hypothetical reversible pathway<br />

where ΔG = w useful ), then the expression<br />

q P = TΔS (ΔG = q P + w useful –TΔS)<br />

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140


Thus q P , which is pathway-dependent, varies<br />

between ΔH (when w useful = 0) & TΔS (w useful =<br />

w useful<br />

max<br />

).<br />

TΔS represents the minimum heat flow that must<br />

accompany the process under consideration. That<br />

is, TΔS represents the minimum energy that must<br />

be “wasted” through heat flow as the process<br />

occurs.<br />

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In summary - at constant T & P<br />

(I) q P = ΔH if w useful = 0<br />

(II) q P = TΔS if w useful = w useful<br />

max<br />

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142


10.14 Adiabatic processes<br />

Adiabatic process (v.s. isothermal process) : a<br />

process in which no energy as heat flows into or<br />

out of the system.<br />

For an adiabatic process<br />

q = 0 & ΔE = q + w = w<br />

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Figure 10.18 : An ideal gas<br />

confined in an isolated<br />

container with a movable<br />

piston. No heat flow with<br />

the surroundings can occur.<br />

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In an adiabatic expansion (v.s. in an isothermal<br />

process) the temperature of the gas decreases (the<br />

average kinetic energy of sample decreases) to<br />

furnish the energy to do the work.<br />

E = nC v Τ (ideal gas, whose energy is dependent<br />

on temp)<br />

ΔE = w = -P ex ΔV= nC v ΔT<br />

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For an infinitesimal adiabatic change<br />

dE = -P ex dV = nC v dT<br />

For a reversible process (P ex ~ P gas )<br />

P ex = P gas = nRT/V<br />

Thus, for a reversible, adiabatic expansioncompression<br />

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

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-(nRT/V)dV = nC v dT<br />

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

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Ideal gas law<br />

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°C<br />

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For a reversible isothermal expansion at 298 K from<br />

P 1 = 10.0 atm & V 1 = 12.2 L to P 2 = 1.0 atm, the final<br />

volume is<br />

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Figure 10.19<br />

Comparison of the<br />

adiabatic &<br />

isothermal<br />

expansion for ideal<br />

samples in which n<br />

= 5 & P 1 = 10.0 atm<br />

& P 1 = 12.2 L.<br />

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For a reversible, isothermal expansion :<br />

P 1 V 1 = P 2 V 2<br />

or PV = constant<br />

For a reversible, adiabatic expansion :<br />

P 1 V 1γ =P 2 V 2γ<br />

or PV γ = constant<br />

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155

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