Kelter P., Mosher M., Scott A. Chemistry. The Practical Science

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which energy and matter can be distributed also increases. Proba-

bility predicts that if the multiplicity of the system increases, there

should be a corresponding increase in the number of ways in which

energy and matter can be distributed in the system. This growth in

the number of microstates increases the entropy of a system. In

other words, the more probable outcome of a spontaneous process is

that an increase in entropy occurs.

We can perform a simple experiment to help explain entropy.

Say we have a friend place a bottle of perfume at one end of a room

while we sit blindfolded in a chair on the opposite side of the room,

as illustrated in Figure 14.11. Our friend opens the bottle. Still

blindfolded, can we tell that the bottle has been opened? At ﬁrst we

would say it is still closed. Given a little time, though, we begin to

notice the fragrance of the perfume and conclude that the bottle

was opened. Why do we smell the perfume? The fragrance was

released from the bottle on the opposite side of the room, so

shouldn’t it remain near the opened bottle? You know from experi-

ence that this isn’t the case. What causes the perfume molecules to

diffuse throughout the room and, eventually, into the noses of peo-

ple at its most distant points? There is no pressure difference across the room, but

still the perfume mixes spontaneously with the air. We can assume from the ki-

netic molecular theory that the attractive forces between the molecules of per-

fume and of those in the air are negligible, so there should be no signiﬁcant

change in the enthalpy of the process. Diffusion is neither exothermic nor en-

dothermic for ideal gases. Instead, diffusion increases the distribution of the mol-

ecules throughout the room. Diffusion also increases the distribution of energy.

The entropy of the system has increased.

We need to be extremely careful when we think of entropy. In processes repre-

sented by the perfume experiment, it appears that the level of disorder of the

molecules has increased. Sometimes, we incorrectly think of entropy as a measure

of the disorder of a system. But even though the increase in entropy often paral-

lels the increase in disorder, entropy is not a measure of how disorganized a

system has become. Disorder is a macroscopic description of a system, whereas

entropy is related to the number of microstates.

The Second Law

Inside the cells of your body lie the enzymes (polymers of amino acids; see Chap-

ter 12) that release energy from glucose. As shown in Figure 14.12, these large

polymers start out as long ﬂexible strands but, shortly after being made, fold into

a small globular shape that contains pockets for binding glucose, ATP, ADP, water,

and other compounds. The structure and type of amino acids around the bind-

ing pockets determine what type of reaction the enzyme will catalyze. Protein

folding into the correct shape is a spontaneous process.

Does this make sense? The

ﬂexible extended chain has many more motions than the folded enzyme; the

number of ways to distribute energy in the system decreases as the enzyme folds.

On the basis of this information alone, we might predict that protein folding is

nonspontaneous. A closer look reveals our need for a deeper understanding of

entropy and spontaneity.

As a rule, we say that a spontaneous process is accompanied by an increase in

the entropy of the universe. This is the

second law of thermodynamics. Mathemat-

ically, the change in the entropy of the universe is greater than zero for a sponta-

neous process:

∆S

universe

> 0

588 Chapter 14 Thermodynamics: A Look at Why Reactions Happen

Increase in entropy

FIGURE 14.11

An experiment to explain entropy. Diffusion of gases is

driven by entropy.

Application

∆S

universe

Is the Sum of ∆S

system

and ∆S

surroundings

∆S

surroundings

∆S

system

∆S

universe

Spontaneity

+++Spontaneous

+−? Spontaneous if ∆S

system

< ∆S

surroundings

−+? Spontaneous if ∆S

system

> ∆S

surroundings

−−−Nonspontaneous

TABLE 14.2

Water

Protein

Nonpolar

group

FIGURE 14.12

The folding of proteins is driven by

entropy. The unfolded protein (a) disrupts

the interactions of the water mole-

cules. The nonpolar groups are tucked

inside the protein, removing them from

interaction with the solvent (b). The fold-

ing also increases the intermolecular

forces of attraction between different

regions of the protein and increases the

number of interactions between solvent

molecules.

However, recall the ﬁrst law of thermodynamics (from Chapter 5), which says

that the energy of the universe is constant (∆E

universe

= 0). This contrasts with the

second law, which implies that the entropy of the universe constantly increases. In

other words, the number of possibilities for the distribution of the energy and

matter of the universe constantly increases. This increase is related to the entropy

changes in the system and surroundings (see Chapter 5 if you wish to review our

deﬁnitions of the terms universe, system, and surroundings). Because the total en-

tropy of the universe is the sum of the change in entropy for a particular system

(∆S

system

) and the change in entropy of the surroundings (∆S

surroundings

), we can

describe the change in entropy of the universe as follows:

∆S

universe

= ∆S

system

+ ∆S

surroundings

■

If ∆S

universe

> 0, the process is spontaneous.

■

If ∆S

universe

< 0, the process is nonspontaneous and is the reverse of the spon-

taneous process.

■

If ∆S

universe

is zero, we say that the process is neither spontaneous nor

nonspontaneous but is at equilibrium, a condition of energetic stability that

we will discuss shortly.

Because we take the sum of the change in entropy of the system and the change

in entropy of the surroundings to obtain the change in entropy of the universe,

∆S

system

could be a negative number and the overall process could still remain

spontaneous. For example, if ∆S

system

= –50 J/K·mol and ∆S

surroundings

=+80 J/

mol·K, then

∆S

system

+ ∆S

surroundings

= ∆S

universe

−50 J/mol·K + (+80 J/mol·K) =+30 J/mol·K

In this case, ∆S

surroundings

increases more than ∆S

system

decreases, so ∆S

universe

in-

creases and the process is spontaneous. Table 14.2 outlines the effects of ∆S

universe

as a function of the change in entropy of the system and surroundings. Pick some

14.2 Why Do Chemical Reactions Happen? Entropy and the Second Law of Thermodynamics 589

(a) (b)

Visualization: Spontaneous

Reactions

590 Chapter 14 Thermodynamics: A Look at Why Reactions Happen

The protein resides

in a “pocket” of space.

The folded protein has

smaller disruptions of

solvent–solvent interactions.

Water

FIGURE 14.13

The unfolded protein reduces the inter-

actions of the solvent molecules. By fold-

ing, it allows those favorable interac-

tions to take place.

sample values for ∆S

system

and ∆S

surroundings

, as we did just above, to help clarify

the outcomes in the table.

In the case of our folding protein, we must take into account the entropy of

the system and that of the surroundings if we are to properly assess the change in

entropy of the universe and determine whether the process is spontaneous. As an

extended chain, the enzyme’s nonpolar groups must interact with the aqueous

cellular environment—the surroundings—as shown in Figure 14.13. The un-

folded enzyme disrupts many of the interactions that occur among the solvent

molecules (water). To reduce those disruptions within the surroundings, the

folding enzyme tucks its nonpolar groups into the center of the globular struc-

ture, effectively removing their interaction with the water inside the cell. The

folding allows the polar groups in the enzyme to interact with the polar water

molecules in the surroundings. The water molecules are also able to interact with

each other with minimal disruption. The number of microstates for the protein

decreases during the process of folding, but the number of microstates for the

water and the water–protein interaction increases. The result is a

S

surroundings

that is more positive than the negative S

system

. And the overall process, the fold-

ing of a protein, is spontaneous (

S > 0).

EXERCISE 14.3 Reaction Spontaneity

Determine whether the values shown below will produce a spontaneous process.

S

system

= 140 J/mol·K; S

surroundings

= –155 J/mol·K

Solution

In a spontaneous process the entropy of the universe increases, so we must add the

values for the system and surroundings to answer the question.

S

universe

= S

system

+ S

surroundings

= 140 J/mol·K +(−155 J/mol·K)

=−15 J/mol·K

Because our calculated value for the entropy of the universe is negative, the process

is nonspontaneous.

PRACTICE 14.3

Determine whether the values for entropy in each of these cases will produce a

spontaneous process.

a. S

system

=−23 J/mol·K; S

surroundings

= –55 J/mol·K

b. S

system

= 38 J/mol·K; S

surroundings

= 59 J/mol·K

c. S

system

=−84 J/mol·K; S

surroundings

= 132 J/mol·K

See Problems 17, 18, 29, and 30.

14.3 Temperature and Spontaneous Processes

Living organisms have been found all over the Earth, from the mouths of near-

boiling geysers to the depths of the Arctic Ocean. These living organisms, just like

the species that live near the tropics, require energy to survive. Many of them

use glucose as a source of energy. In fact, some of the psychrotrophs (bacteria

that can survive exposure to low temperatures) and psychrophiles (bacteria that

thrive in low temperatures) employ a concentrated solution of glucose as an

“antifreeze” because such a solution lowers their freezing point. Research by

microbiologists and biochemists into the life processes of the psychrophiles indi-

cates that these organisms, like a lot of other living things on the planet, break

down glucose to produce energy through glycolysis. Does the temperature at

which these organisms live affect the spontaneity of the reactions involved in

glycolysis? To answer this question, we need to consider the signs on the change

in entropy of the system (∆S

system

) and of the surroundings (∆S

surroundings

). A

positive change in the entropy of the universe is required for the process to be

spontaneous.

In general, as compounds undergo changes in their physical states from solid

to liquid to gas, the entropy of the system increases (∆S

system

=+), as shown in

Figure 14.16. Water molecules in an ice cube have a well-deﬁned order in an ice

cube. As the ice begins to melt, this well-deﬁned order disappears and the water

molecules increase their range of motions and, therefore, the number of mi-

crostates. The number of microstates that are possible in the liquid water suggests

that the melting of ice corresponds to an increase in the entropy of the system.

The same holds true when water is con-

verted into steam. Conversely, the reverse

of these processes (gas to liquid to solid)

is typically accompanied by a loss of

entropy (∆S

system

= –).

What is the effect of energy transfer

on ∆S? When steam condenses to liquid

water, energy as heat ﬂows out of the sys-

tem and into the surroundings, and the

kinetic energy of the particles in the sur-

roundings increases. The motions of the

atoms in the surroundings increase, and

the sign of ∆S

surroundings

is positive. On

the other hand, if energy as heat ﬂows

from the surroundings to the system

(liquid to vapor), we’d expect the kinetic

14.3 Temperature and Spontaneous Processes 591

Application

HEMICAL ENCOUNTERS:

Psychrotrophs and

Psychrophiles

Water molecules gain increased

freedom of motion in changing

phase from solid to liquid.

Video Lesson: Entropy and

Temperature

592 Chapter 14 Thermodynamics: A Look at Why Reactions Happen

Imagine living at the bottom of the Arctic Ocean, think-

ing life was grand inside glacial ice, enjoying the weather

near a thermal vent, or relaxing under the crush of 1 mile

of bedrock. These conditions sound fairly extreme to us,

but not to a class of bacteria known as the extremophiles.

Some of these microorganisms thrive near the hot bub-

bling mud-pots of Yellowstone National Park, others in

the sulfur-laden waters near a geothermal vent at the bot-

tom of the Atlantic Ocean. Extremophiles, some examples

of which are listed in Table 14.3, survive in conditions

that we humans would ﬁnd extreme.

Biochemists, microbiologists, and geologists from

around the world study these creatures because of the ex-

treme conditions in which they live and to learn more

about the enzymes that continue to work under the

equally extreme conditions inside them. For instance,

millions of Americans are lactose-intolerant and have

difﬁculty digesting the lactose in almost every dairy prod-

uct, including milk and ice cream. Imagine if you could

isolate beta-galactosidase (an enzyme that breaks down

sugars like lactose into more easily digested compounds)

from an extremophile that lived in icy environments. The

extremophile’s beta-galactosidase should be capable of

working quite well in cold environments. By adding the

isolated beta-galactosidase to milk and related products

like ice cream, you could make lactose-free dairy products

without having to heat them. Researchers at Pennsylvania

State University have been able to show that this is possi-

ble by isolating a strain of bacteria, known as Arthrobacter

psychrolactophilus that has a modiﬁed beta-galactosidase

that works best when the temperature is 15°C and contin-

ues to work well when the temperature is as low as 0°C.

Thomas D. Brock isolated the ﬁrst example of a true

extremophile from hot springs, such as that shown in

Classes of Extremophiles

Class Extreme Environment Locations Where They Live

Acidophiles Low pH Sulfurous springs and acid mine drainage

Alkaliphiles High pH Alkaline lakes and basic soils

Anaerobes Non-oxygen-containing environments Fermenting juices

Barophiles High pressure Deep sea vents and deep within the Earth

Copiotrophs High nutrient levels Sugar solutions

Halophiles High ion concentration Saline lakes and salt deposits

Hyperthermophiles Temperatures above 70°C Hydrothermal vents, hot springs

Methanogens Methane-rich Deep-sea vents, oil deposits

Oligotrophs Low nutrient levels Desert, rocks

Psychrophiles Low temperatures, typically below 10°C Glaciers, Arctic Ocean, cold soils

Thermophiles Temperatures above 50°C Hydrothermal vents, hot springs

TABLE 14.3

NanoWorld / MacroWorld

Big effects of the very small:

Industrial uses for the extremophiles

solid

< <<<S

liquid

gas

FIGURE 14.16

Entropy increases as a compound changes

state from solid to liquid to gas. Note that

the increased molecular motions allow more

microstates to exist in the compound.

14.3 Temperature and Spontaneous Processes 593

animal as an aid in digesting food. Their use would im-

prove the usefulness of cheap food as a source of energy

for the animals. The alkaliphiles (base-tolerant bacteria)

thrive in basic soils such as those in the western United

States and in Egypt. Proteases (enzymes that break down

proteins) and lipases (enzymes that break down oils) iso-

lated from these bacteria could ﬁnd potential use in the

detergent industry. Their addition to laundry detergents

(which typically are basic) would improve the ability of

the detergent to clean stains from clothing.

Investigators are currently searching, and ﬁnding,

bacteria that live in environments we originally thought

were sterile. After determining the types and properties

of the enzymes that these bacteria possess, scientists are

exploiting their industrial utility. The uses of these en-

zymes as catalysts to aid human life are endless. Will we

ﬁnd extremophiles that proliferate on Mars? . . . on Io, the

volcanic innermost major moon of Jupiter? . . . on our

own Moon? And, if so, what uses might we ﬁnd for the

enzymes they produce?

FIGURE 14.14

The Morning Glory Pool atYellowstone National Park is named for the

brightly colored thermophiles that ﬂourish in this high-temperature

environment.

FIGURE 14.15

Hydrothermal vent in the North Paciﬁc west of Vancouver Island

on the Juan de Fuca Ridge. Extremophiles such as Pyrolobus

fumarii and Methanopyrus live on the sides of these chimneys.

The “smoke” ﬂowing from the vents is actually made up of

minerals from the lava under the ocean ﬂoor.

Figure 14.14, in Yellowstone National Park in Wyoming.

This bacterium, called Thermus aquaticus, grows most

rapidly at temperatures near 70°C. Other thermophiles,

or heat-loving bacteria, include Sulfolobus acidocaldarius,

which lives in sulfur-laden hot springs at temperatures as

high as 85°C, and Pyrolobus fumarii, which is isolated

from deep-sea hydrothermal vents (Figure 14.15), grows

only at temperatures above 90°C, and reproduces best at

105°C). These bacteria are of industrial interest because

of their ability to grow at such high temperatures. For ex-

ample, the enzyme Taq polymerase (isolated from T.

aquaticus) is used in DNA ﬁngerprinting because it can

survive the severe temperature variations in the poly-

merase chain reaction used to make multiple copies of

puriﬁed DNA (see Chapter 22).

The acid-tolerant extremophiles (acidophiles) are of

interest because their enzymes are capable of operating in

highly acidic environments. A potential application for

the acidophile’s enzymes is their addition to cattle feed,

because they would work well in the acidic gut of an

energy of the particles in the surroundings to decrease and the motions of the

atoms (and, therefore, the entropy) in the surroundings to decrease also. In short,

a ﬂow of energy as heat out of the system and into the surroundings (an exothermic

process) corresponds to a positive sign for ∆S

surroundings

. An endothermic process

has an opposite effect on the surroundings; endothermic processes correspond

to a negative sign for ∆S

surroundings

What is this exchange of energy to which we refer? If our process occurs under

reversible conditions at a constant pressure, we can relate the energy of the

process (q

rev

) to the change in enthalpy of the process (∆H). Reversible conditions

occur when the process is allowed to proceed in inﬁnitesimally small steps. At any

point during the reaction, we could change the direction of the reaction with

Energy

System

Surroundings

Exothermic processes involve the

transfer of energy from the system

to the surroundings.

merely slight modiﬁcations. Often, reversible conditions exist during phase changes.

Quantitatively, we can summarize our statements by saying that the change in the

entropy is equal to the change in enthalpy of the phase change (a reversible

process) divided by the temperature:

S

system

rev

H

system

where the temperature (T) is reported in kelvins (K) and the enthalpy (∆H

system

)

is reported in joules per mole. Because of some assumptions we’ve made to arrive

at this equation, its use is limited to describing heat transfers when the tempera-

ture remains constant. For example, the equation works well for describing phase

transitions but poorly for describing a reaction.

Because q

surroundings

=–q

system

, and q

system

=∆H

system

, the value of ∆S

surroundings

can be obtained using a similar equation.

S

surroundings

−H

system

EXERCISE 14.4 Entropy Change at a Phase Change

Instead of carrying water in their backpacks, hikers can melt ice to make drinking

water. They

also boil water for drinking and food preparation. What is the en-

tropy change of the system for melting 1 mol of ice at 0.00°C and 1 atm? What

is the entropy change for melting 125 g of ice? The enthalpy of fusion at this

phase change,



fus

H is 6.01 kJ/mol.

O(s) n H

O(l) 

fus

H = 6.01 kJ/mol at 0.0°C

First Thoughts

The entropy change for the system or surroundings can be calculated if we know the

temperature and the enthalpy of the system. Because the units of entropy are usu-

ally reported in J/mol·K, we should convert the units for the enthalpy and the tem-

perature to match, so our calculation is simpliﬁed. The second part of the question

asks us to examine the speciﬁc entropy change for a quantity of water that is not

equal to 1 mol. We may do this part of the calculation by dimensional analysis.

Solution

The change in the entropy of this reversible process can be calculated using the

formula we just discussed. This process involves an increase in the entropy of

the system.

S

system

H

system

6010 J/mol

273 K

= 22.0 J/mol·K

We’ve calculated the change in entropy (S = 22.0 J/mol·K) for the

phase change, so we can use this value to determine the change in en-

tropy for melting 125 g of water.

S

system

= 125 g ×

1 mol

18.02 g

22.0J

mol·K

153 J

Further Insights

Reactions can occur with either an increase or a decrease of entropy.

It is interesting to note that this change is based on the number of

moles of the compounds in the reaction. If this relationship appears

594 Chapter 14 Thermodynamics: A Look at Why Reactions Happen

Melting ice into water.

to be similar to what we observed in our discussion of enthalpy in Chapter 5, it

should. Also, as we’ll see later, entropy can be manipulated in ways that are very sim-

ilar to those we used for enthalpy.

Moreover, because the temperature remains constant, we can calculate the value

of S

surroundings

. We ﬁnd that the value is –22.0 J/mol·K. Therefore, the entropy

change for the universe (S

universe

) should be 0.0 J/mol·K (S

universe

= S

system

S

surroundings

). This process is neither spontaneous nor nonspontaneous. It is

reversible.

PRACTICE 14.4

Calculate S

system

for each of these processes at 25°C. Assume that each is a re-

versible process.

(g) n I

(s) 

sub

H =+62.4 kJ

O(l) n H

O(g) 

vap

H =+40.7 kJ

See Problems 35 and 36.

HERE’S WHAT WE KNOW SO FAR

■

Spontaneous processes are associated with an increase in the entropy of the

universe.

■

The rate of a spontaneous reaction is not related to its spontaneity.

■

The reverse of a spontaneous process is a nonspontaneous process.

■

The change in entropy of the universe can be calculated as the sum of the

change in entropy of the system and the change in entropy of the sur-

roundings.

■

The entropy of the system can be calculated for reversible processes by divid-

ing the enthalpy for the process by the temperature of the process.

14.4 Calculating Entropy Changes

in Chemical Reactions

Metabolism (the biochemical reactions of an organism) releases energy. In this

process, the potential energy stored in food is used by an organism, and some of

the food is converted via chemical reactions into molecules that are needed for

the organism to survive. All of these reactions are spontaneous in the body and

vital to the living processes that occur at the cellular level. For instance, the body

breaks down sucrose, perhaps contained in a sugar-laden gumdrop, into glucose

and fructose. The glucose then enters a series of reactions, the glycolytic pathway

being the ﬁrst (Figure 14.9), as the body converts it into carbon dioxide and

water. Along the way it produces ATP (Figure 14.10), a molecule used to store po-

tential energy. Plants, on the other hand, synthesize glucose by combining carbon

dioxide and water. This reaction is energetically uphill for the plant, so it uses the

high-energy ATP molecule to drive the reaction to completion. Why is the break-

down of glucose energetically downhill, whereas the formation of glucose is up-

hill? In other words, why does it take energy to make glucose, and why is energy

released when glucose is broken down? We can answer this question by examin-

ing the entropy changes in the combustion of glucose. Qualitatively, is the sign of

14.4 Calculating Entropy Changes in Chemical Reactions 595

Application

HEMICAL

ENCOUNTERS:

More Glycolysis

596 Chapter 14 Thermodynamics: A Look at Why Reactions Happen

∆S

system

positive or negative? The reactants include 1 mol of glucose molecules

and 6 mol of gaseous oxygen molecules.

(s) + 6O

(g) n 6O

(g) + 6H

O(g)

The combustion proceeds as 7 mol of reactants are converted into 12 mol of

products (6 mol of gaseous carbon dioxide and 6 mol of gaseous water). Prior to

the reaction, there were a large number of possible microstates because of the

large number (7 mol) of reactants. Because gases occupy a larger volume in a

ﬂask than do equivalent quantities of solids, the 6 mol of oxygen in our reaction

occupy the majority of the locations within the ﬂask as they rapidly travel within

it. After the reaction, there are 12 mol of gas inside the ﬂask. The number of mi-

crostates increased because of the larger number of gaseous products. The in-

crease in the number of microstates is an increase in the entropy of the system

(∆S

system

is positive), as shown in Figure 14.17. If we examine the reverse of this

reaction (from the viewpoint of the glucose-producing plant), the entropy of the

system is lowered because we are combining 12 mol of gaseous carbon dioxide

and water to make 6 mol of gaseous oxygen and 1 mol of solid sugar.

What we’ve discussed is a method by which one can usually predict the sign of

the entropy change accurately. As a rule, the change in entropy of a reaction is pos-

itive if the number of gaseous molecules increases. Although the number of moles of

solid and liquid molecules contributes to the overall number of microstates, the

large volume occupied by gaseous molecules contributes much more. By simply

examining a reaction, we can predict the entropy change. The change is much

harder to assess for reactions that do not involve gaseous molecules.

The reaction of glucose with oxygen

generates carbon dioxide and water.

Compare the number of molecules of

gaseous products made to the number

of molecules of gaseous reactants.

(s) + 6O

(g) 6CO

(g) + 6H

O(g)

→

reactants

< S

products

FIGURE 14.17

An increase in the number of gaseous

molecules increases the entropy of the

system.

EXERCISE 14.5 Predict the Sign

1. Some camp stoves use butane as fuel. The combustion of butane is exothermic,

providing heat to cook food and boil water. Predict the sign of S

system

for the

combustion of butane.

2CH

(g) + 13O

(g) n 8CO

(g) + 10H

O(g)

2. The Haber process, the combination of hydrogen and nitrogen gases to form

ammonia gas (NH

) is one of the most widely used manufacturing processes

because of worldwide demand for ammonia-based fertilizer. Predict the sign

of S

system

for the production of ammonia.

(g) + N

(g) n 2NH

(g)

3. Barium hydroxide (Ba(OH)

·8H

O) reacts with ammonium chloride (NH

Cl)

to form several products, including ammonia, water, and barium chloride.

Predict the sign of ∆S

system

for this reaction.

Ba(OH)

·8H

O(s) + 2NH

Cl(s) n BaCl

(aq) + 2NH

(g) + 10H

O(l)

Solution

1. There are 18 mol of gaseous products and only 15 mol of gaseous reactants.

Consequently, the number of microstates should increase for the reaction.

This leads us to predict that the change in entropy should be a positive

number for this reaction. Based on calculations that we’ll discover later,

S

system

=+789 J/mol·K. This agrees with our prediction.

2. The Haber process takes 4 mol of gaseous reactants and forms 2 mol of gaseous

products. The entropy change of the system is likely to be negative for this re-

action. The actual ∆S

system

value is –199 J/mol·K.

3. The reaction results in the formation of 2 mol of ammonia gas for each mole

of barium hydroxide octahydrate that reacts with solid ammonium chloride.

Consequently, we would predict that the change in system entropy would be

positive. One factor to keep in mind is that the waters of hydration (that is, the

eight H

O molecules that are part of the barium hydroxide crystal) would be

released as part of the process. These molecules would tend to raise the sys-

tem entropy further as they join the liquid state. The actual ∆S

system

value is

468 J/mol·K.

PRACTICE 14.5

Predict the sign of the change in entropy for each of these reactions.

O(l) n H

O(g)

OH(l) + HCl(g) n CH

Cl(l) + H

O(l)

See Problems 39–42, 45, and 47.

Qualitatively, our prediction of the sign of ∆S

system

can be based on the num-

ber of gaseous molecules in the reaction. Quantitatively, the value for the change

in the entropy (∆S

system

) is nearly as easy to determine. However, a problem

arises. In order to calculate a change in a state function, we subtract the ﬁnal value

from the initial value. How do we determine the initial value of entropy for a

compound?

14.4 Calculating Entropy Changes in Chemical Reactions 597

Butane