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11 Fossil Fuels, Fuel Efficiency, and Heats of Combustion

11 Fossil Fuels, Fuel Efficiency, and Heats of Combustion

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290



Chapter 8 THERMOCHEMISTRY: CHEMICAL ENERGY



Coal is burned just as it comes from the mine, but petroleum must be refined

before use. Refining begins with distillation, the separation of crude liquid oil into

fractions on the basis of their boiling points (bp). So-called straight-run gasoline

(bp 30–200 °C) consists of compounds with 5–11 carbon atoms per molecule;

kerosene (bp 175–300 °C) contains compounds in the C11 -C14 range; gas oil

(bp 275–400 °C) contains C14 -C25 substances; and lubricating oils contain whatever

remaining compounds will distill. Left over is a tarry residue of asphalt (Figure 8.11).



Figure 8.11



The products of petroleum refining.



Asphalt



Lubricating oil

Waxes

Gas oil

(C14−C25)

Petroleum

Kerosene

(C11−C14)



The different fractions

are grouped according

to the number of carbon

atoms their molecules

contain.



Straight-run gasoline

(C5−C11)

Natural gas

(C1−C4)



As the world’s petroleum deposits become more scarce, other sources of energy

will have to be found to replace them. Hydrogen, although it burns cleanly and is relatively nonpolluting, has numerous drawbacks: low availability, difficulty in

transport and storage, and low combustion enthalpy per milliliter, to name a few.

Ethanol and methanol look like potential choices for alternative fuels because both

can be produced relatively cheaply and have reasonable combustion enthalpies. At

present, ethanol is produced largely by fermentation of corn or cane sugar, but

methods are being developed to produce it from waste wood by the breakdown of

cellulose to glucose and subsequent fermentation. Methanol is produced directly

from natural gas in a two-step process:

CH 4(g) + H 2O(g) ¡ CO(g) + 3 H 2(g)

CO(g) + 2 H 2(g) ¡ CH 3OH(l)

Ī PROBLEM 8.19



Butane



Liquid butane (C4H10), the fuel used in many disposable lighters, has

¢H°f = -147.5 kJ/mol and a density of 0.579 g/mL. Write a balanced equation for the

combustion of butane, and use Hess’s law to calculate the enthalpy of combustion in

kJ/mol, kJ/g, and kJ/mL.



8.12 AN INTRODUCTION TO ENTROPY



291



8.12 AN INTRODUCTION TO ENTROPY

We said in the introduction to this chapter that chemical reactions (and physical

processes) occur when the final state is more stable than the initial state. Because less

stable substances generally have higher internal energy and are converted into more

stable substances with lower internal energy, energy is generally released in chemical

reactions. At the same time, though, we’ve seen that some reactions and processes

occur even though they absorb rather than release energy. The endothermic reaction

of barium hydroxide octahydrate with ammonium chloride shown previously in

Figure 8.6, for example, absorbs 80.3 kJ of heat (¢H° = +80.3 kJ) and leaves the surroundings so cold that the temperature drops below 0 °C.

Ba(OH)2 # 8 H 2O(s) + 2 NH 4Cl(s) ¡ BaCl2(aq) + 2 NH 3(aq) + 10 H 2O(l)

¢H° = +80.3 kJ

An example of a physical process that takes place spontaneously yet absorbs

energy takes place every time an ice cube melts. At a temperature of 0 °C, ice spontaneously absorbs heat from the surroundings to turn from solid into liquid water.

What’s going on? Because the reaction of barium hydroxide octahydrate with

ammonium chloride and the melting ice cube absorb heat yet still take place spontaneously, there must be some other factor in addition to energy that determines

whether a reaction or process will occur. We’ll take only a brief look at this additional

factor now and return for a more in-depth study in Chapter 16.

Before exploring the situation further, it’s important to understand what the

word spontaneous means in chemistry, for it’s not quite the same as in everyday language. In chemistry, a spontaneous process is one that, once started, proceeds on its

own without a continuous external influence. The change need not happen quickly,

like a spring uncoiling or a sled going downhill. It can also happen slowly, like the

gradual rusting away of an iron bridge or abandoned car. A nonspontaneous process,

by contrast, takes place only in the presence of a continuous external influence.

Energy must be continuously expended to re-coil a spring or to push a sled uphill.

When the external influence stops, the process also stops.

What do the reaction of barium hydroxide octahydrate and the melting of an ice

cube have in common that allows the two processes to take place spontaneously

even though they absorb heat? The common feature of these and all other processes

that absorb heat yet occur spontaneously is an increase in the amount of molecular

randomness of the system. The eight water molecules rigidly held in the

Ba(OH)2 # 8 H 2O crystal break loose and become free to move about randomly in the

aqueous liquid product. Similarly, the rigidly held H2O molecules in the ice lose their

crystalline ordering and move around freely in liquid water.

The amount of molecular randomness in a system is called the system’s entropy

(S). Entropy has the units J/K (not kJ/K) and is a quantity that can be determined for

pure substances, as we’ll see in Section 16.5. The larger the value of S, the greater the

molecular randomness of the particles in the system. Gases, for example, have more

randomness and higher entropy than liquids, and liquids have more randomness

and higher entropy than solids (Figure 8.12).

A change in entropy is represented as ¢S = Sfinal - Sinitial. When randomness

increases, as it does when barium hydroxide octahydrate reacts or ice melts, ¢S has a

positive value because Sfinal 7 Sinitial. The reaction of Ba(OH)2 # 8 H2O(s) with

NH4Cl(s) has ¢S° = +428 J/K, and the melting of ice has ¢S° = +22.0 J/(K # mol).

When randomness decreases, ¢S is negative because Sfinal 6 Sinitial. The freezing of

water, for example, has ¢S° = -22.0 J/(K # mol). (As with ¢H°, the superscript ° is

used in ¢S° to refer to the standard entropy change in a reaction where products and

reactants are in their standard states.)

Thus, two factors determine the spontaneity of a chemical or physical change in a

system: a release or absorption of heat (¢H) and an increase or decrease in molecular



᭡ Sledding downhill is a spontaneous

process that, once started, continues on its

own. Dragging the sled back uphill is a

nonspontaneous process that requires a

continuous input of energy.



292



Chapter 8 THERMOCHEMISTRY: CHEMICAL ENERGY



Figure 8.12



Entropy is a measure of molecular

randomness.



Liquids have more

randomness and higher

entropy than solids.



Gases have more

randomness and higher

entropy than liquids.



Less randomness,

lower entropy



More randomness,

higher entropy



randomness (¢S). To decide whether a process is spontaneous, both enthalpy and

entropy changes must be taken into account:

Spontaneous process:



Favored by decrease in H (negative ¢H)

Favored by increase in S (positive ¢S)



Nonspontaneous process:



Favored by increase in H (positive ¢H)

Favored by decrease in S (negative ¢S)



Note that the two factors don’t have to operate in the same direction. Thus, it’s

possible for a process to be disfavored by enthalpy (endothermic, positive ¢H) yet still

be spontaneous because it is strongly favored by entropy (positive ¢S). The melting

of ice [ ¢H° = +6.01 kJ/mol; ¢S° = +22.0 J/(K # mol)] is just such a process, as

is the reaction of barium hydroxide octahydrate with ammonium chloride

(¢H° = +80.3 kJ; ¢S° = +428 J/K). In the latter case, 3 mol of solid reactants

produce 10 mol of liquid water, 2 mol of dissolved ammonia, and 3 mol of dissolved

ions (1 mol of Ba2+ and 2 mol of Cl-), with a consequent large increase in molecular

randomness:

Ba(OH)2 · 8 H2O(s) + 2 NH4Cl(s)

3 mol solid reactants



BaCl2(aq) + 2 NH3(aq) + 10 H2O(l)

3 mol

2 mol dissolved

dissolved ions

molecules



10 mol

liquid

water molecules



ΔH° = +80.3 kJ



Unfavorable



ΔS° = +428 J/K



Favorable



Conversely, it’s also possible for a process to be favored by enthalpy (exothermic,

negative ¢H) yet be nonspontaneous because it is strongly disfavored by entropy

(negative ¢S). The conversion of liquid water to ice is nonspontaneous above 0 °C,



8.13 AN INTRODUCTION TO FREE ENERGY



for example, because the process is disfavored by entropy [ ¢S° = -22.0 J/(K # mol)]

even though it is favored by enthalpy (¢H° = -6.01 kJ/mol).

WORKED EXAMPLE 8.11



PREDICTING THE SIGN OF ¢S FOR A REACTION

Predict whether ¢S° is likely to be positive or negative for each of the following

reactions:

(a) H 2C “ CH 2(g) + Br2(g) ¡ BrCH 2CH 2Br(l)

(b) 2 C2H 6(g) + 7 O 2(g) ¡ 4 CO 2(g) + 6 H 2O(g)

STRATEGY



Look at each reaction, and try to decide whether molecular randomness increases or

decreases. Reactions that increase the number of gaseous molecules generally have a

positive ¢S, while reactions that decrease the number of gaseous molecules have a

negative ¢S.

SOLUTION



(a) The amount of molecular randomness in the system decreases when 2 mol of

gaseous reactants combine to give 1 mol of liquid product, so the reaction has a

negative ¢S°.

(b) The amount of molecular randomness in the system increases when 9 mol of gaseous

reactants give 10 mol of gaseous products, so the reaction has a positive ¢S°.

Ī PROBLEM 8.20 Ethane, C2H6, can be prepared by the reaction of acetylene, C2H2,

with hydrogen. Is ¢S° for the reaction likely to be positive or negative? Explain.



C2H 2(g) + 2 H 2(g) ¡ C2H 6(g)

CONCEPTUAL PROBLEM 8.21 Is the reaction represented in the following drawing

likely to have a positive or a negative value of ¢S°? Explain.



Reaction



8.13 AN INTRODUCTION TO FREE ENERGY

How do we weigh the relative contributions of enthalpy changes (¢H) and entropy

changes (¢S) to the overall spontaneity of a process? To take both factors into

account when deciding the spontaneity of a chemical reaction or other process, we

define a quantity called the Gibbs free-energy change (≤G), which is related to ¢H

and ¢S by the equation ¢G = ¢H - T¢S.

Free-energy

change



Heat of

reaction



Temperature

(in kelvin)



Entropy

change



ΔG = ΔH − TΔS



The value of the free-energy change ¢G determines whether a chemical or physical process will occur spontaneously. If ¢G has a negative value, free energy is

released and the process is spontaneous. If ¢G has a value of 0, the process is neither

spontaneous nor nonspontaneous but is instead at an equilibrium. And if ¢G has a

positive value, free energy is absorbed and the process is nonspontaneous.

≤G<0

≤G ‫ ؍‬0

≤G>0



Process is spontaneous

Process is at equilibrium—neither spontaneous nor nonspontaneous

Process is nonspontaneous



293



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Chapter 8 THERMOCHEMISTRY: CHEMICAL ENERGY



The fact that the T¢S term in the free-energy equation is temperature dependent

implies that some processes might be either spontaneous or nonspontaneous,

depending on the temperature. At low temperature, for instance, an unfavorable

(positive) ¢H term might be larger than a favorable (positive) T¢S term, but at

higher temperature, the T¢S term might be larger. Thus, an endothermic process that

is nonspontaneous at low temperature can become spontaneous at higher temperature. This, in fact, is exactly what happens in the ice/water transition. At a

temperature below 0 °C, the melting of ice is nonspontaneous because the unfavorable ¢H term outweighs the favorable T¢S term. At a temperature above 0 °C,

however, the melting of ice is spontaneous because the favorable T¢S term outweighs the unfavorable ¢H term (Figure 8.13). At exactly 0 °C, the two terms are

balanced.

¢G° = ¢H° - T¢S°

At -10 °C (263 K):



¢G° = 6.01



kJ

kJ

- (263 K)a0.0220 #

b = +0.22 kJ/mol

mol

K mol



At 0 °C (273 K):



¢G° = 6.01



kJ

kJ

- (273 K)a0.0220 #

b = 0.00 kJ/mol

mol

K mol



At +10 °C (283 K):



¢G° = 6.01



kJ

kJ

- (283 K)a0.0220 #

b = -0.22 kJ/mol

mol

K mol



Figure 8.13



Melting and freezing. The

melting of ice is disfavored by

enthalpy (¢H 7 0) but favored

by entropy (¢S 7 0). The

freezing of water is favored

by enthalpy (¢H 6 0) but

disfavored by entropy (¢S 6 0).



ΔS° = +22.0 J/(K · mol)

(Entropy increases)

ΔH° = +6.01 kJ/mol

(Endothermic)

Spontaneous above 0 °C

Spontaneous below 0 °C

ΔS° = −22.0 J/(K · mol)

(Entropy decreases)

ΔH° = −6.01 kJ/mol

(Exothermic)



Solid water

Below 0 °C, the enthalpy

term ΔH dominates the

entropy term TΔS in the

Gibbs free-energy equation,

so freezing is spontaneous.



Liquid water

At 0 °C the entropy

and enthalpy terms

are exactly balanced.



Above 0 °C, the

entropy term

dominates the

enthalpy term,

so melting is

spontaneous.



An example of a chemical reaction in which temperature controls spontaneity is

that of carbon with water to yield carbon monoxide and hydrogen. The reaction has

an unfavorable ¢H term (positive) but a favorable T¢S term (positive) because randomness increases when a solid and 1 mol of gas are converted into 2 mol of gas:

C(s) + H 2O(g) ¡ CO(g) + H 2(g)



¢H° = +131 kJ

¢S° = +134 J/K



Unfavorable

Favorable



No reaction occurs if carbon and water are mixed at room temperature because

the unfavorable ¢H term outweighs the favorable T¢S term. At approximately

978 K (705 °C), however, the reaction becomes spontaneous because the favorable

T¢S term becomes larger than the unfavorable ¢H term. Below 978 K, ¢G has a positive value; at 978 K, ¢G = 0; and above 978 K, ¢G has a negative value. (The

calculation is not exact because values of ¢H and ¢S themselves vary somewhat

with temperature.)



8.13 AN INTRODUCTION TO FREE ENERGY



¢G° = ¢H° - T¢S°

kJ

b = +1 kJ

K

kJ

¢G° = 131 kJ - (978 K)a0.134

b = 0 kJ

K

kJ

¢G° = 131 kJ - (988 K)a0.134 b = -1 kJ

K

¢G° = 131 kJ - (968 K)a0.134



At 695 °C (968 K):

At 705 °C (978 K):

At 715 °C (988 K):



The reaction of carbon with water is, in fact, the first step of an industrial process

for manufacturing methanol (CH3OH). As supplies of natural gas and oil diminish,

this reaction may become important for the manufacture of synthetic fuels.

A process is at equilibrium when it is balanced between spontaneous and nonspontaneous—that is, when ¢G = 0 and it is energetically unfavorable to go either

from reactants to products or from products to reactants. Thus, at the equilibrium

point, we can set up the equation

¢G = ¢H - T¢S = 0 At equilibrium

Solving this equation for T gives

¢H

T =

¢S

which makes it possible to calculate the temperature at which a changeover in

behavior between spontaneous and nonspontaneous occurs. Using the known values of ¢H° and ¢S° for the melting of ice, for instance, we find that the point at

which liquid water and solid ice are in equilibrium is

¢H°

=

¢S°



6.01 kJ

= 273 K = 0 °C

kJ

0.0220

K

Not surprisingly, the ice/water equilibrium point is 273 K, or 0 °C, the melting point

of ice.

In the same way, the temperature at which the reaction of carbon with water

changes between spontaneous and nonspontaneous is 978 K, or 705 °C:

131 kJ

¢H°

= 978 K

T =

=

kJ

¢S°

0.134

K

This section and the preceding one serve only as an introduction to entropy and

free energy. We’ll return in Chapter 16 for a more in-depth look at these two important topics.

T =



WORKED EXAMPLE 8.12



USING THE FREE-ENERGY EQUATION TO CALCULATE

EQUILIBRIUM TEMPERATURE

Lime (CaO) is produced by heating limestone (CaCO3) to drive off CO2 gas, a reaction

used to make Portland cement. Is the reaction spontaneous under standard conditions

at 25 °C? Calculate the temperature at which the reaction becomes spontaneous.

CaCO 3(s) ¡ CaO(s) + CO 2(g) ¢H° = 179.2 kJ; ¢S° = 160.0 J/K

STRATEGY



The spontaneity of the reaction at a given temperature can be found by determining

whether ¢G is positive or negative at that temperature. The changeover point between

spontaneous and nonspontaneous can be found by setting ¢G = 0 and solving for T.

SOLUTION



At 25 °C (298 K), we have

kJ

b = +131.5 kJ

K

Because ¢G is positive at this temperature, the reaction is nonspontaneous.

¢G = ¢H - T¢S = 179.2 kJ - (298 K)a0.1600



continued on next page



295



296



Chapter 8 THERMOCHEMISTRY: CHEMICAL ENERGY



The changeover point between spontaneous and nonspontaneous is approximately

T =



179.2 kJ

¢H

=

= 1120 K

¢S

kJ

0.1600

K



The reaction becomes spontaneous above approximately 1120 K (847 °C).



WORKED CONCEPTUAL EXAMPLE 8.13



PREDICTING THE SIGNS OF ¢H, ¢S, AND ¢G FOR A REACTION

What are the signs of ¢H, ¢S, and ¢G for the following nonspontaneous transformation?



STRATEGY



First, decide what kind of process is represented in the drawing. Then decide whether

the process increases or decreases the entropy of the system and whether it is exothermic or endothermic.

SOLUTION



The drawing shows ordered particles in a solid subliming to give a gas. Formation of a

gas from a solid increases molecular randomness, so ¢S is positive. Furthermore,

because we’re told that the process is nonspontaneous, ¢G is also positive. Because the

process is favored by ¢S (positive) yet still nonspontaneous, ¢H must be unfavorable

(positive). This makes sense, because conversion of a solid to a liquid or gas requires

energy and is always endothermic.



Ī PROBLEM 8.22



Which of the following reactions are spontaneous under standard

conditions at 25 °C, and which are nonspontaneous?

¢G° = -55.7 kJ

(a) AgNO 3(aq) + NaCl(aq) ¡ AgCl(s) + NaNO 3(aq)

(b) 2 C(s) + 2 H 2(g) ¡ C2H 4(g)



¢G° = 68.1 kJ



Ī PROBLEM 8.23 Is the Haber process for the industrial synthesis of ammonia spontaneous or nonspontaneous under standard conditions at 25 °C? At what temperature (°C)

does the changeover occur?



N2(g) + 3 H 2(g) ¡ 2 NH 3(g)

CONCEPTUAL PROBLEM 8.24



¢H° = -92.2 kJ; ¢S° = - 199 J/K



The following reaction is exothermic:



(a) Write a balanced equation for the reaction.

(b) What are the signs of ¢H and ¢S for the reaction?

(c) Is the reaction likely to be spontaneous at low temperatures only, at high temperatures only, or at all temperatures? Explain.



INQUIRY WHAT ARE BIOFUELS?



297



INQUIRY WHAT ARE BIOFUELS?

The petroleum era began in August 1859, when the world’s first oil well was drilled

near Titusville, Pennsylvania. Since that time, approximately 1.2 * 1012 barrels of

petroleum have been used throughout the world, primarily as fuel for automobiles

(1 barrel = 42 gallons).

No one really knows how much petroleum remains on Earth. Current world consumption is approximately 3.1 * 1010 barrels per year, and currently known

recoverable reserves are estimated at 1.1 * 1012 barrels. Thus, the world’s known

petroleum reserves will be exhausted in approximately 35 years at the current rate of

consumption. Additional petroleum reserves will surely be found, but consumption

is also increasing, making any prediction of the amount of time remaining highly

inaccurate. Only two things are certain: The amount of petroleum remaining is finite,

and we will run out at some point, whenever that might be. Thus, alternative energy

sources are needed.

Of the various alternative energy sources now being explored, biofuels—fuels

derived from recently living organisms such as trees, corn, sugar cane, and

rapeseed—look promising because they are renewable and because they are more

nearly carbon neutral than fossil fuels, meaning that the amount of CO2 released to the

environment during the manufacture and burning of a biofuel is similar to the

amount of CO2 removed from the environment by photosynthesis during the plant’s

growth. Note that the phrase carbon neutral doesn’t mean that biofuels don’t release

CO2 when burned; they release just as much CO2 as any other fuel.

The two biofuels receiving the most attention at present are ethanol and

biodiesel. Ethanol, sometimes called bioethanol to make it sound more attractive, is

simply ethyl alcohol, the same substance found in alcoholic drinks and produced in

the same way: by yeast-catalyzed fermentation of carbohydrate.



H



HO



H



C

H



C



C



H



OH



OH



O

H



C



C



H



OH



Glucose

(C6H12O6)



H



OH

C

H



Yeast enzymes



2



H



C



H



H

C



OH + 2 CO2



H



Ethanol

(C2H5OH)



The only difference between beverage ethanol and fuel ethanol is the source of

the sugar. Beverage ethanol comes primarily from fermentation of sugar in grapes

(for wine) or grains (for distilled liquors), while fuel ethanol comes primarily from

fermentation of cane sugar or corn. Much current work is being done, however, on

developing economical methods of converting cheap cellulose-based agricultural

and logging wastes into fermentable sugars.



᭡ Vegetable oil from the bright yellow

rapeseed plant is a leading candidate for

large-scale production of biodiesel fuel.



298



Chapter 8 THERMOCHEMISTRY: CHEMICAL ENERGY



Biodiesel consists primarily of organic compounds called long-chain methyl esters,

which are produced by reaction of common vegetable oils with methyl alcohol in the

presence of an acid catalyst. Any vegetable oil can be used, but rapeseed oil and soybean oil are the most common. (Canola oil, of which you may have heard, is a specific

cultivar of generic rapeseed.) Once formed, the biodiesel is typically mixed in up

to 30% concentration with petroleum-based diesel fuel for use in automobiles

and trucks.



O

H3C



H2

C



H2

C



H2

C

C

H2



C

H2



H2

C

C

H2



H2

C

C

H2



H2

C

C

H2



H2

C

C

H2



H2

C

C

H2



CH3



C

C

H2



O



A typical long-chain methyl ester in biodiesel



Ī PROBLEM 8.25 Write balanced equations for the combustion reactions of ethanol

(C2H6O) and biodiesel (C19H38O2) with oxygen to give CO2 and H2O.

Ī PROBLEM 8.26 Biodiesel has a more favorable (more negative) combustion enthalpy

per gram than ethanol. Explain, in light of your answer to Problem 8.25.



CONCEPTUAL PROBLEMS



299



SUMMARY

Energy is either kinetic or potential. Kinetic energy (EK) is the energy

of motion. Its value depends on both the mass m and velocity v of

an object according to the equation EK = (1/2)mv 2. Potential

energy (EP) is the energy stored in an object because of its position

or in a chemical substance because of its composition. Heat is the

thermal energy transferred between two objects as the result of a

temperature difference, whereas temperature is a measure of the

kinetic energy of molecular motion.

According to the conservation of energy law, also known as

the first law of thermodynamics, energy can be neither created

nor destroyed. Thus, the total energy of an isolated system is constant. The total internal energy (E) of a system—the sum of all

kinetic and potential energies for each particle in the system—is a

state function because its value depends only on the present condition of the system, not on how that condition was reached.

Work (w) is defined as the distance moved times the force that

produces the motion. In chemistry, most work is expansion work

(PV work) done as the result of a volume change during a reaction

when air molecules are pushed aside. The amount of work done by

an expanding gas is given by the equation w = -P¢V, where P is

the pressure against which the system must push and ¢V is the

change in volume of the system.

The total internal energy change that takes place during a reaction is the sum of the heat transferred (q) and the work done

(-P¢V). The equation

¢E = q + (-P¢V)



or



q = ¢E + P¢V = ¢H



smaller than the ¢E term, so that the total internal energy change

of a reacting system is approximately equal to ¢H, also called

the heat of reaction. Reactions that have a negative ¢H are

exothermic because heat is lost by the system, and reactions that

have a positive ¢H are endothermic because heat is absorbed by

the system.

Because enthalpy is a state function, ¢H is the same regardless

of the path taken between reactants and products. Thus, the sum of

the enthalpy changes for the individual steps in a reaction is equal

to the overall enthalpy change for the entire reaction, a relationship

known as Hess’s law. Using this law, it is possible to calculate

overall enthalpy changes for individual steps that can’t be measured directly. Hess’s law also makes it possible to calculate the

enthalpy change of any reaction if the standard heats of formation

(¢H°f) are known for the reactants and products. The standard

heat of formation is the enthalpy change for the hypothetical formation of 1 mol of a substance in its thermodynamic standard

state from the most stable forms of the constituent elements in

their standard states (1 atm pressure and a specified temperature,

usually 25 °C).

In addition to enthalpy, entropy (S)—a measure of the amount

of molecular randomness in a system—is also important in determining whether a process will occur spontaneously. Together,

changes in enthalpy and entropy define a quantity called the

Gibbs free-energy change (¢G) according to the equation

¢G = ¢H - T¢S. If ¢G is negative, the reaction is spontaneous;

if ¢G is positive, the reaction is nonspontaneous.



where ¢H is the enthalpy change of the system, is a fundamental

equation of thermochemistry. In general, the P¢V term is much



KEY WORDS

conservation of energy

law 267

endothermic 277

enthalpy (H) 273

enthalpy change ( ≤H) 273

entropy (S) 291

exothermic 277

first law of

thermodynamics 268



Gibbs free-energy change

( ≤G) 293

heat 268

heat capacity (C) 279

heat of combustion

( ≤H°c) 289

heat of reaction ( ≤H) 273

Hess’s law 282

internal energy (E) 268



molar heat capacity

(Cm) 279

specific heat 279

spontaneous process 291

standard enthalpy of

reaction ( ≤H°) 275

standard heat of formation

( ≤H°f) 284

state function 269



sublimation 276

system, surroundings 268

temperature 268

thermochemistry 267

thermodynamic standard

state 275

work (w) 270



CONCEPTUAL PROBLEMS

Problems 8.1–8.26 appear within the chapter.

8.27 The following reaction is exothermic:



(a) Write a balanced equation for the reaction (red spheres

represent A atoms and ivory spheres represent B

atoms).

(b) What are the signs ( + or - ) of ¢H and ¢S for the

reaction?

(c) Is the reaction likely to be spontaneous at lower temperatures only, at higher temperatures only, or at all

temperatures?



300



Chapter 8 THERMOCHEMISTRY: CHEMICAL ENERGY



8.28 Imagine a reaction that results in a change in both volume

and temperature:

1 atm

1 atm



Reaction



V=2L

T = 300 K



V=3L

T = 310 K



(1) CH 3CH 2OH(l) + 1/2 O2(g) ¡

CH 3CHO(g) + H 2O(l)



¢H° = -174.2 kJ



(2) CH 3CHO(g) + 1/2 O2(g) ¡

CH 3CO2H(l)



¢H° = -318.4 kJ



Net CH 3CH 2OH(l) + O2(g) ¡

CH 3CO2H(l) + H 2O(l)



¢H° = -492.6 kJ



8.32 A reaction is carried out in a cylinder fitted with a movable

piston. The starting volume is V = 5.00 L, and the apparatus is held at constant temperature and pressure. Assuming

that ¢H = -35.0 kJ and ¢E = -34.8 kJ, redraw the piston

to show its position after reaction. Does V increase,

decrease, or remain the same?

1 atm



(a) Has any work been done? If so, is its sign positive or

negative?

(b) Has there been an enthalpy change? If so, what is the

sign of ¢H? Is the reaction exothermic or endothermic?

8.29 Redraw the following diagram to represent the situation

(a) when work has been gained by the system and (b) when

work has been lost by the system:

1 atm



V = 5.00 L



8.33 The following drawing portrays a reaction of the type

A ¡ B + C, where the different colored spheres represent different molecular structures. Assume that the

reaction has ¢H° = + 55 kJ. Is the reaction likely to be

spontaneous at all temperatures, nonspontaneous at all

temperatures, or spontaneous at some but nonspontaneous

at others? Explain.

Reaction

Reaction



8.30 Acetylene, C2H2, reacts with H2 in two steps to yield

ethane, CH3CH3:

(1)

HC CH + H2

(2) H2C CH2 + H2

Net HC CH + 2 H2



H2C CH2 ΔH° = −175.1 kJ

CH3CH3

ΔH° = −136.3 kJ

CH3CH3



8.34 What are the signs of ¢H, ¢S, and ¢G for the following

spontaneous change? Explain.



ΔH° = −311.4 kJ



(a)

(c)

Hess’s law diagram



8.35 The following reaction of A3 molecules is spontaneous:



(b)





Which arrow (a, b, c) in the Hess’s law diagram corresponds to which step, and which arrow corresponds to the

net reaction? Where are the reactants located on the diagram, and where are the products located?

8.31 Draw a Hess’s law diagram similar to the one in Problem

8.30 for the reaction of ethyl alcohol (CH3CH2OH) with

oxygen to yield acetic acid (CH3CO2H).



(a) Write a balanced equation for the reaction.

(b) What are the signs of ¢H, ¢S, and ¢G for the reaction?

Explain.



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