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

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188 Chapter 5 Energy

Water as gas, liquid,

and solid.

This is an exothermic reaction.

Many different standard enthalpies of reactions have been deﬁned. Common

ones include the standard enthalpy of formation and the standard enthalpy of

combustion.The

standard enthalpy of formation, 

H° (also known as the standard

heat of formation

) of a substance is the enthalpy change for the formation of

1 mole of the substance in its standard state from its elements in their reference

forms. For example, the standard enthalpy of formation of carbon dioxide from

carbon and oxygen in their reference forms, which for carbon is graphite, can be

summarized as follows:

C(s) + O

(g) → CO

(g) 

H° =−394 kJ/mol

We have to show just 1 mole of the product, because that value is embodied in the

deﬁnition. This can force us to use fractional amounts of moles of some of the

reactants, as in the next example, which indicates the standard enthalpy of for-

mation of water:

(g) +

(g) → H

O(l) 

H° =−286 kJ/mol

We must show the oxygen in its standard state as O

(g). We cannot simply use O

(because O is not the reference form of oxygen), so we must indicate half a mole

of O

. Keep in mind that when you are writing standard enthalpy equations,

it is acceptable to leave equations with fractional stoichiometric values, rather

than multiplying by some common factor to convert all the fractions to whole

numbers.

EXERCISE 5.6 To the Moon

The combustion of gaseous hydrazine (N

) is a reaction that provides thrust for

rockets. This reaction is also used on the space shuttle for minor attitude changes

when the shuttle is in orbit. Write the equation that describes the standard enthalpy

of formation for this compound. Given that for hydrazine 

H° =+95.40 kJ/mol,

calculate the change in enthalpy accompanying the formation of 125 g of hydrazine

from its elements in their reference forms.

First Thoughts

The equation that we prepare needs to describe the formation of just 1 mole of hy-

drazine. Because hydrazine contains nitrogen and hydrogen, we can show its for-

mation from those elements in their reference forms. We can then convert the num-

ber of grams into moles and use that value to determine the enthalpy for the

reaction.

5.4 Enthalpy 189

Solution

The equation for the formation of gaseous hydrazine from its elements in their ref-

erence forms is

The number of moles of hydrazine can then be used to calculate the enthalpy.

125 g N

1 mol N

32.05 g N

= 3.90 mol N

3.90 mol N

95.40 kJ

1 mol N

= 372 kJ

Further Insights

The amounts of materials that we use in reactions determine the change in en-

thalpy of the reaction. Conversely, the enthalpy change in a reaction can be used to

determine the amount of material consumed or produced in the reaction.

PRACTICE 5.6

Calculate the enthalpy change for the formation of 38.0 g of water. (Use the appen-

dix to ﬁnd the standard enthalpy change for water.)

See Problems 47–50.

Elements in their reference forms are the basic starting materials for all the re-

actions associated with standard enthalpies of formation. The elements themselves

are not formed from any simpler chemicals, so the standard enthalpies of formation

of elements in their reference forms are all equal to zero. This is to say that, using

calcium as an example,

Ca(s) → Ca(s) 

H° = 0 kJ/mol

Note that 

H° = 0 kJ/mol does not mean that U = 0 kJ/mol. Also, recall that

only one form of an element is its reference form. For example, O

(g) is the refer-

ence form for oxygen. When O

forms the allotrope ozone, O

, in the atmosphere,



H° is not equal to 0 kJ/mol.

(g) → O

(g) 

H° = 143 kJ/mol

The standard enthalpy of combustion, 

H° (also known as the standard heat of

combustion

) of a substance is the enthalpy change when 1 mole of the substance

in its standard state is completely burned in oxygen gas. For example, consider

the combustion of propane gas (C

), which we ﬁrst discussed in Exercise 5.5.

(g) + 5O

(g) → 3CO

(g) + 4H

O(l) 

H° =−2202 kJ/mol

(g) + N

(g)N

(g)

→

When we write the equation to illustrate the standard enthalpy of combustion,

we must show exactly 1 mole of the reactant, propane, being burned. The coefﬁ-

cients must be adjusted to ﬁt that requirement, even as the mole ratios of all reac-

tants and products necessarily remain the same. Consider, as well, the equation

In some cases, we will ﬁnd it useful to multiply all of the equation coefﬁcients

by some number. Let’s see what happens when we multiply the coefﬁcients in the

“combustion of carbon” equation by 2:

C(s) + O

(g) → CO

(g) + 394 kJ/mol

2C(s) + 2O

(g) → 2CO

(g) + 2 × (394 kJ/mol)

Note that although the mole ratios of the reactants and products remain the

same, the enthalpy change is affected by this multiplication. Remembering that

enthalpy as a product indicates a negative change in enthalpy, we note that H of

this reaction is now (2

×−394 kJ/mol) =−788 kJ/mol.

This ability to reverse equations and automatically know the corresponding

enthalpy change is a consequence of what we know as Hess’s law, which is the

subject of the next section. It enables us to determine the enthalpy changes of

reactions that might be very difﬁcult to actually perform.

EXERCISE 5.7 How Much Heat?

A camper wishes to warm a pot of water on a propane stove. She calculates that

she’ll need to produce 209 kJ to increase the temperature of her pot of water by

50°C. How much propane, in moles, will be consumed to produce this much heat?

(g) + 5O

(g) → 3CO

(g) + 4H

O(l) 

H° =−2202 kJ/mol

190 Chapter 5 Energy

FIGURE 5.18

Negative enthalpy changes can be

thought of as adding enthalpy to the

product side of the equation. Reversing

the direction of the equation changes

the sign of the enthalpy change.

C(s) + O

(g) → CO

(g) + 394 kJ/mol 

H° =−394 kJ/mol

394 kJ/mol + CO

(g) → C(s) + O

(g) H =+394 kJ/mol

for the standard enthalpy of combustion of carbon. Note that the equation is the

same as the equation for the standard enthalpy of formation of carbon dioxide.

C(s) + O

(g) → CO

(g) 

H° =−394 kJ/mol = 

H°(CO

)

Manipulating Enthalpies

When we know the value for any standard enthalpy change of a reaction, we also

automatically know the value for the enthalpy change of the reverse reaction. For

example, converting carbon dioxide gas back into solid carbon and oxygen gas

would be the reverse of the forward reaction, the combustion of carbon to form

. It should therefore be accompanied by an enthalpy change that is equal to

that of the forward reaction but opposite in sign, as shown in Figure 5.18. To help

us understand what the signs on the enthalpy terms mean, we can consider this

manipulation as though the enthalpy change were part of the equation. Let’s

examine the following equations and see how this applies to the end result. The

enthalpy of combustion is negative, so it must be released from the reaction and,

hence, appear on the same side as the products in this exothermic process.

C(s) + O

(g) → CO

(g) + 394 kJ/mol

The reverse of this reaction is

(g) + 394 kJ/mol → C(s) + O

(g)

which shows that the enthalpy is gained by the reaction.

First Thoughts

We must incorporate the enthalpy of combustion into the equation and determine

the stoichiometric relationship between moles of propane (C

(g)) and the

change in enthalpy. Accordingly, we can write

1 mol C

(g) =−2202 kJ

1 mol C

2202 kJ

Solution

209 kJ ×

1 mol C

2202 kJ

= 0.0949 mol C

(g )

Note that we drop the negative sign from the enthalpy change because the amount

of heat released from the system is the amount of heat needed to warm the water.

Further Insights

The thermodynamic sign conventions can be a little misleading and confusing at

ﬁrst. In the present example, the quantity of enthalpy released (209 kJ) by the com-

bustion reaction (the system) is equal to the quantity of enthalpy gained by the sur-

roundings (209 kJ), assumed here to be the pot of water, so although the quantities

are equal, the signs are opposite: H

system

=−209 kJ; H

surround

=+209 kJ.

PRACTICE 5.7

Calculate how many moles of the compound printed in boldface type will be

needed to produce a change in enthalpy of 250 kJ for each of these reactions.

a. 2NH

(g) + 3N

O(g) → 4N

(g) + 3H

O(l) H =−1012 kJ

b. 2N

O(g) → O

(g) + 2N

(g) H =−164 kJ

See Problems 34, 35, 71b, 71c, 72b, 72c, 76, and 99b.

5.5 Hess’s Law

The search for alternative fuel sources to supplement what we currently use is a

high priority in all ﬁelds of transportation, including commercial aviation, per-

sonal automobile use, and even rocketry. The search for such fuels could involve

the examination of lots of new reactions, some of which might never have been

performed before. Fortunately, one of the features of enthalpy change values

works to our beneﬁt.We can calculate the energy or enthalpy changes of reactions

without having to perform them. The reason for this ease of calculation lies in the

fact that energy (U) and enthalpy (H) are state functions. A

state function is a

property of a system that depends only on its present state, not on how it got there.

State functions are usually represented by an italic capital letter.

For an illustration of this idea, consider the different paths that the space

shuttle might take to reach the international space station, orbiting at an altitude

of 240 miles above sea level, as graphically shown in Figure 5.19. One shuttle mis-

sion may be launched directly up to a 240-mile-high orbit, ready to rendezvous

with the space station. A later shuttle mission might ﬁrst go into a higher orbit to

release a satellite and then come back down to the 240-mile-high orbit of the

space station. The eventual altitude of each shuttle is the same: 240 miles above

sea level. In short, the ﬁnal altitude is independent of the path taken. The dis-

tances traveled by the two shuttles in reaching their ﬁnal altitude are very

different, however, because one took the direct route and the other took a more

circuitous route. Altitude is a state function; distance traveled is not.

5.5 Hess’s Law 191

Video Lesson: Hess’s Law

Video Lesson: Enthalpies of

Formation

Tutorial: Hess’s Law

192 Chapter 5 Energy

Higher

orbit

route

Direct

route

FIGURE 5.19

Two shuttles visiting the International

Space Station will end up at the same al-

titude, regardless of the route taken. In

this example, altitude is a state function;

its value depends only on the initial and

ﬁnal states, not on the route taken.

C then:

(b)

(a)

∆H

FIGURE 5.20

A diagrammatic summary of Hess’s law. (a) If the distance down the

mountain from A to B and the distance up to the top from A to C are

known, the distance from B to C can be determined. (b) The enthalpy

change for a series of reactions is analogous to this. If the enthalpy of

reaction from A to B and from A to C is known, the enthalpy of reac-

tion from B to C can be calculated.

Products

Progress of reaction

Energy

Reactants

100 J

Forward

reaction

releases

100 J

Reverse

reaction

absorbs

100 J

FIGURE 5.21

A forward reaction has an enthalpy that is equal, but op-

posite in sign, to that of its reverse reaction.

The change in the value of any state function depends only on the initial and

ﬁnal states between which it is changing. In the case of enthalpy,

H = H

ﬁnal

− H

initial

If a chemical reaction is carried out by two different chemical paths, the overall

enthalpy change associated with each path will be the same. That basic principle

is known as

Hess’s law and is summarized diagrammatically in Figure 5.20.

Enthalpy is a state function, which also justiﬁes our earlier statement that the

enthalpy change for any reaction will have the same value as the enthalpy change

for the reverse reaction but will be opposite in sign. If the forward reaction has an

enthalpy change of

−100 joules, the reverse reaction will have an enthalpy change

+100 joules. This is summarized diagrammatically in Figure 5.21.

Visualization: Hess’s Law

Using Hess’s Law

On the Salt River-Pima Maricopa Indian Community in Arizona, methane (CH

)

collected from a 10-million-ton garbage landﬁll is used to fuel a“green” (environ-

mentally benign) power plant that provides electricity for 2000 homes. Methane

is a chemically very simple and convenient fuel, arising from the biological degra-

dation of foodstuffs and other wastes in the landﬁll. It also makes up about 95%

of the “natural gas” found above oil deposits, is burned to provide energy for

domestic cooking and heating, and is as a source of heat in many industries.

Table 5.3 and Figure 5.22 list the important sources of methane emis-

sions into the atmosphere. The data in the table suggest that it is not pos-

sible to accurately quantify the production of a gas that is emitted every

day from so many sources. The numbers should be seen as rough projec-

tions of quantities that cannot be measured with great certainty. Nonethe-

less, harvesting some of this methane, as at the Salt River Project in Ari-

zona, could represent an important addition to the world’s energy supply.

Although methane found in natural gas is formed in a variety of dif-

ferent biological and chemical ways, it can be described by the following

chemical equation:

C(s) + 2H

(g) → CH

(g)

If we use Hess’s law, we can calculate the standard enthalpy change for this

reaction without actually performing the reaction. To complete the calcu-

lation, we need to know the standard enthalpy changes of other reactions

that will allow us to consider forming methane indirectly. We use reference

tables to obtain these reactions and their corresponding standard en-

thalpies. These values are

C(s) + O

(g) → CO

(g) H° =−394 kJ/mol

(g) +

(g) → H

O(l) H° =−286 kJ/mol

(g) + 2O

(g) → CO

(g) + 2H

O(l) H° =−890 kJ/mol

5.5 Hess’s Law 193

Projected U.S. Methane Emissions in the Year 2020

Source Teragrams (1 Tg = 10

g) of CH

Landﬁlls 7.0

Coal mining 5.5

Natural gas/oil drilling 7.0

Livestock manure 4.5

Livestock digestion releases (“enteric fermentation”) 6.5

Other 0.5

TOTAL 31.0

TABLE 5.3

Methane gas can be harvested from landﬁlls.

Application

HEMICAL

ENCOUNTERS:

Focus on Methane

Natural

occurrences

Anthropogenic

influences

Ruminants

Termites

Wetlands

Ocean water

Fresh water

Rice paddies

Biomass

burning

Landfills

Coal

mining

Gas

production

Methane

hydrate

FIGURE 5.22

Major sources of atmospheric methane as reported by the

U.S. Department of Energy. Sources can be broken down into

those that occur naturally and those that occur through an-

thropogenic inﬂuences (that is, through human activity).

194 Chapter 5 Energy

890 kJ/mol

(g) + 2O

(g)

–76 kJ/mol

–394 kJ/mol

–286 kJ/mol

C(s) + O

(g)

(g) + O

(g)

O(l )

(g) + O

(g)

–

FIGURE 5.23

A diagrammatic representation of how

Hess’s law can be used to calculate the

enthalpy of formation of methane.

We arrange these reactions in such a way that they allow us to calculate the enthalpy

of formation of methane. This is shown diagrammatically in Figure 5.23. Alterna-

tively, we can rearrange the chemical reactions to mathematically arrive at the

same answer. The underlying logic to Hess’s law calculations is that we can mul-

tiply, and/or reverse, and then combine any chemical equations in whatever way

will achieve the overall chemical change in which we are interested. For example,

to calculate the enthalpy of formation of methane, we can manipulate the three

combustion equations as follows:

■

C(s) + O

(g) → CO

(g) H° =−394 kJ/mol

■

2 × [H

(g) +

(g) → H

O(l)] 2 × [H° =−286 kJ/mol]

= 2H

(g) + O

(g) → 2H

O(l) =−572 kJ

■

−1 × [CH

(g) + 2O

(g) → CO

(g) + 2H

O(l)] −1 × [H° =−890 kJ/mol]

= CO

(g) + 2H

O(l) → CH

(g) + 2O

(g) =+890 kJ

Adding the three equations together, we calculate the overall enthalpy change:

C(s) + O

(g) → CO

(g) H° =−394 kJ

(g) + O

(g) → 2H

O(l) H° =−572 kJ

(g) + 2H

O(l) → CH

(g) + 2O

(g) H° =+890 kJ

C + O

+ 2H

+ O

+ CO

+ 2H

O → CO

+ 2H

O + CH

+ 2O

H° =−76 kJ

When we cancel the chemicals that appear in equal amounts on both sides of the

equation (O

,CO

O), this simpliﬁes to the equation for the formation of

methane:

C + O

+ 2H

+ O

+ CO

+ 2H

O → CO

+ 2H

O + CH

+ 2O

C(s) + 2H

(g) → CH

(g)

H° =−76 kJ/mol

Whichever way we decide to calculate the change in enthalpy for the reaction

of interest, we should arrive at the same correct answer. In either case, we don’t

have to perform the ﬁnal reaction to be able to calculate its enthalpy change.

EXERCISE 5.8 Hess’s Law

Given the following data, calculate the value of H for the reaction shown below.

(s) + 10Cl

(g) → 4PCl

(g) H =−2139 kJ

PCl

(g) + Cl

(g) → PCl

(g) H =−155 kJ

(s) + 6Cl

(g) → 4PCl

(g) H = ?? kJ

Solution

1 × [P

(s) + 10 Cl

(g) → 4PCl

(g)] 1 × H =−2139 kJ

−4 × [PCl

(g) + Cl

(g) → PCl

(g)] −4 × H =+620 kJ

(s) + 6Cl

(g) → 4PCl

(g) H =−2139 kJ + 620 kJ =−1519 kJ

PRACTICE 5.8

Given the following data, calculate the value of H for the reaction shown below.

6Fe(s) + 4O

(g) → 2Fe

(s) H =−1787 kJ

2Fe

(s) +

(g) → 3Fe

(s) H =−186 kJ

3Fe

(s) → 6Fe(s) +

(g) H = ?? kJ

See Problems 75–78, 81, and 82.

Reaction Enthalpies from Enthalpies of Formation

The energy needed to maneuver the space shuttle when it is in orbit is provided

by the reaction between methylhydrazine (CH

NHNH

) and dinitrogen tetrox-

ide (N

4CH

NHNH

(l) + 5N

(l) → 4CO

(g) + 9N

(g) + 12H

O(l)

One of the requirements for a chemical reaction that can be used as part of an

effective propulsion system is that it have a large negative H value, indicating

that it releases a lot of energy. How can we determine the value of H for this

reaction?

One answer, based on our earlier discussion, is to perform the reaction in a

calorimeter—something that we may not wish to do, especially if we are studying

rocket fuel. Alternatively, we could use Hess’s law to calculate the enthalpy of the

reaction. However, it is often difﬁcult to ﬁnd all of the enthalpies that are needed

to complete the circuitous route from reactants to products. Instead, there is an-

other way to determine the enthalpy of the reaction. We can calculate H of the

reaction by means of a useful general rule about enthalpies of formation if other

appropriate values of H are known:

The standard enthalpy change for a reaction can be calculated by subtracting

the sum of the enthalpies of formation of the reactants from the sum of the

enthalpies of formation of the products.

This can be expressed mathematically as follows:

H° reaction =

n



◦

(products) −

n



◦

(reactants)

where the Greek symbol

 (sigma) means “the sum of the following terms.”

= the number of moles of each product in the reaction equation

= the number of moles of each reactant in the reaction equation

5.5 Hess’s Law 195

Methylhydrazine

196 Chapter 5 Energy

4CH

NHNH

(l)

4C(s)

(g)

12H

(g)

4CO

(g)

12H

O(l)

10O

(g)

(l)

∆H = 4 × (–55 kJ/mol)

= –220 kJ/mol

∆H = 4 × (–394 kJ/mol)

= –1576 kJ/mol

∆H = 12 × (–286 kJ/mol)

= –3432 kJ/mol

∆H = 0.0 kJ/mol

∆H = 5 × (20.0 kJ/mol)

= 100 kJ/mol

∆H = –220 kJ/mol + 100 kJ/mol

= –120 kJ/mol

∆H = –120 kJ/mol + –5008 kJ/mol

= –5128 kJ/mol

Reactants to elements

∆H = –1576 kJ/mol + 0.0 kJ/mol + –3432 kJ/mol

= –5008 kJ/mol

Elements to products

Reactants to products

FIGURE 5.24

We can consider performing a reaction

by taking the indirect route, via the ele-

ments in their reference forms. Then, by

Hess’s law, the reactions shown here can

allow us to determine the enthalpy of

the reaction.

Figure 5.24 clariﬁes why this procedure works.



H°(CO

(g)) =−394 kJ/mol 

H°(N

(g)) = 0.0 kJ/mol



H°(H

O(l)) =−286 kJ/mol

n



◦

(products) = [4 ×−394] + [9 × 0.0] + [12 ×−286] =−5008 kJ



H°(CH

NHNH

(l)) =+55 kJ/mol 

H°(N

(g)) =−20.0 kJ/mol

n



H°(reactants) = [4 ×+55] + [5 ×−20.0] =+120 kJ

H°reaction =

n



◦

(products) –

n



◦

(reactants)

= (−5008 kJ) − (120 kJ) =−5128 kJ

We ﬁnd, as expected, that this reaction used to provide propulsion to maneuver

the space shuttle in orbit has a relatively large negative H value equal to

−5128 kJ/mol.

EXERCISE 5.9 Thermite in Space

One reaction that may prove useful for the welding and brazing that are necessary

when building platforms in space is the “thermite reaction” between aluminum and

iron oxide:

2Al(s) + Fe

(s) → 2Fe(s) +Al

(s)

Calculate the standard enthalpy change of this reaction, given the following

enthalpies of formation:



H°(Fe

) =−824 kJ/mol; 

H°(Al

) =−1676 kJ/mol

First Thoughts

Why don’t reference tables provide heat of formation values for aluminum and

iron? They are in their reference forms, and by deﬁnition, their heat of formation is

zero. This means that the enthalpy change of the reaction (remember, it’s a state

The thermite reaction in action. This

photo shows Austrian railroad workers

in 1910 using the thermite reaction.

The molten iron produced by the reaction

drips between the sections of rail and

welds the track together.

function!) is equal to the heat of formation of the aluminum oxide (Al

) minus

the heat of formation of the iron oxide (Fe

Solution

H°reaction =

n



◦

(products) − n



H°(reactants)

H°reaction = (−1676 + 0) − (−824 + 0) =−852 kJ/mol

Further Insights

In “manned” space ﬂight, many other, less dramatic reactions occur some exother-

mic and some endothermic. In particular, the reaction of molecular hydrogen and

oxygen gases on board the space shuttle to form water and usable electricity is

exothermic, whereas many of the scientiﬁc experiments, such as growing plants,

require the input of energy, rather than its release, to run.

PRACTICE 5.9

Calculate the enthalpy change for the reaction

4NH

(g) + 3O

(g) → 2N

(g) + 6H

O(l)

given the following reactions and H values:

a. 2NH

(g) + 3N

O(g) → 4N

(g) + 3H

O(l) H =−1012 kJ

b. 2N

O(g) → O

(g) + 2N

(g) H =−164 kJ

See Problems 75–78, 81, and 82.

5.6 Energy Choices

Humans have always had a variety of energy sources to choose from, and choices

based on energy issues have always been important. In ancient times, people could

keep warm by burning wood, moving south to sunnier climes, or minimizing

energy losses from their bodies by wrapping themselves in animal skins and furs.

Ancient people also learned to make use of wind energy to power great journeys by

boat acrossseas and oceans.Nowadays,the energychoicesfacing us are much more

complex,and the possible effects of making unwise choices are much greater.

The development of modern civilization was powered largely by burning

things to release heat. This heat could keep people warm and could also be used

to boil water, generating the steam to power steam engines and steam turbines.

The engines and turbines were used to power machinery, trains, and ships and

eventually to make electricity. The greatest leaps forward in technol-

ogy came with the discovery of the fossil fuels—coal, oil, and natural

gas—that could be harvested from the Earth in seemingly limitless

quantities and burned with few obvious disadvantages, apart from the

annoyance of a bit of smoke and the loss of lives in the process of

extracting these natural resources. Today, the burning of fossil fuels

provides about 70% of the energy needed to sustain a complex in-

dustrialized country such as the United States, as shown in Figure

5.25. However, we now know that the supplies of fossil fuels are not

limitless and that burning them to provide energy also generates

problems. The most signiﬁcant problem might be the apparent

warming of the planet, the “global warming” effect, caused in part by

the carbon dioxide gas that is a by-product of burning these fuels.

Because fossil reserves are ﬁnite, the burning of these fuels causes

environmental problems, and their processing and distribution have

had a signiﬁcant impact on the world’s economic and political

5.6 Energy Choices 197

Application

HEMICAL ENCOUNTERS:

Energy Choices

Natural gas,

22.9%

Coal,

22.9%

Geothermal,

solar, wind,

0.5%

Alcohol,

2.8%

Hydropower,

2.7%

Nuclear,

8.2%

Oil,

39.9%

FIGURE 5.25

U.S. Energy use in 2004.

Smog is a serious urban pollutant in

almost every major city in the world.

Video Lesson: CIA

Demonstration: The Thermite

Reaction

Visualization: Thermite Reaction