Chemistry

5. Repeat Preparation steps 3 and 4 for the second CLEANUP AND DISPOSAL
pipet. Label this pipet “Base.” Do not use this
pipet for the acid solution. 5. Clean all equipment and your lab
station. Dispose of chemicals and solu-
6. Make sure that the three trials produce data that tions as directed by your teacher. Wash
are similar to one another. If one is greatly dif- your hands thoroughly before you leave the lab.
ferent from the others, perform Preparation
steps 3–5 again. ANALYSIS AND INTERPRETATION

7. Remove the beaker containing the eggshell from 1. Organizing Ideas: The calcium carbonate in the
the oven. Cool them in a desiccator. Record the eggshell sample undergoes a double-displacement
mass of the entire eggshell in the second table. reaction with the HCl in step 3. Write a balanced
Place half of the shell into the clean mortar, and chemical equation for this reaction. (Hint: The gas
grind the shell into a very fine powder. evolved was CO2.)

PROCEDURE 2. Organizing Ideas: Write the balanced chemical
equation for the acid/base neutralization of the
1. Measure the mass of a piece of weighing paper. excess unreacted HCl with the NaOH.
Transfer about 0.1 g of ground eggshell to a piece
of weighing paper, and measure the eggshell’s 3. Organizing Data: Calculate the volume of each
mass as accurately as possible. Record the mass drop in milliliters. Then convert the number of
in the second data table. Place this eggshell drops of HCl into volume in milliliters. Record
sample into a clean, 50 mL micro solution bottle this volume in your data table. Repeat this step
(or Erlenmeyer flask). for the drops of NaOH.

2. Fill the acid pipet with 1.00 M HCl acid solution, 4. Organizing Data: Using the relationship between
and then empty the pipet into an extra 100 mL the molarity and volume of acid and the molarity
beaker. Label the beaker “Waste.” Fill the base and volume of base needed to neutralize it,
pipet with the 1.00 M NaOH base solution, and calculate the number of moles of the HCl solu-
then empty the pipet into the waste beaker. tion that was neutralized by the NaOH, and
record it in your table. (Hint: This relationship
3. Fill the acid pipet once more with 1.00 M HCl. was discussed in Section 2.)
Holding the acid pipet vertically, add exactly
150 drops of 1.00 M HCl to the bottle or flask 5. Analyzing Results: Calculate the number of
that contains the eggshell. Swirl the flask gently moles of HCl that reacted with the CaCO3 and
for 3 to 4 min. Observe the reaction taking record both in your table.
place. Wash down the sides of the flask with
about 10 mL of distilled water. Using a third CONCLUSIONS
pipet, add two drops of phenolphthalein solu-
tion. 1. Organizing Data: Use the balanced equation for
the reaction to calculate the number of moles of
4. Fill the base pipet with the 1.00 M NaOH. Slowly CaCO3 that reacted with the HCl, and record
add NaOH from the base pipet into the bottle or this number in your table.
flask that contains the eggshell reaction mixture,
counting and recording the drops added. Stop 2. Organizing Data: Use the periodic table to
adding base when the mixture remains a faint calculate the molar mass of CaCO3. Then, use
pink color, even after it is swirled gently. Be sure the number of moles of CaCO3 to calculate the
to add the base drop by drop, and be certain the mass of CaCO3 in your eggshell sample. Record
drops end up in the reaction mixture and not on this mass in your data table. Using the mass of
the walls of the bottle or flask. Record in the sec- CaCO3, calculate the percentage of CaCO3 in
ond data table the number of drops of base used. your eggshell and record it in your data table.

A C I D - B A S E T I T R A T I O N A N D p H 529

CHAPTER 16

Reaction Energy

Many chemical reactions give
off or take in energy as heat.

Kilauea volcano erupting, Hawaii Island

Thermochemistry SECTION 1

V irtually every chemical reaction is accompanied by a change in OBJECTIVES

energy. Chemical reactions usually absorb or release energy as heat. Define temperature and state
You learned in Chapter 10 that energy is also absorbed or released in the units in which it is
physical changes, such as melting a solid or condensing a vapor. measured.
Thermochemistry is the study of the transfers of energy as heat that Define heat and state its units.
accompany chemical reactions and physical changes. Perform specific-heat
calculations.
Heat and Temperature Explain enthalpy change,
enthalpy of reaction, enthalpy
The energy absorbed or released as heat in a chemical or physical change of formation, and enthalpy of
is measured in a calorimeter. In one kind of calorimeter, known quanti- combustion.
ties of reactants are sealed in a reaction chamber, which is immersed in Solve problems involving
a known quantity of water in an insulated vessel. Therefore, the energy enthalpies of reaction,
given off (or absorbed) during the reaction is equal to the energy enthalpies of formation, and
absorbed (or given off) by the known quantity of water. The amount of enthalpies of combustion.
energy is determined from the temperature change of the known mass
of surrounding water. The data collected from calorimetry experiments www.scilinks.org
are temperature changes because energy cannot be measured directly; Topic: Heat/Temperature
but temperature, which is affected by the transfer of energy as heat, is Code: HC60726
directly measurable. To see why this is so, let us look at the definitions
of heat and temperature and at how temperature is measured. R E A C T I O N E N E R G Y 531

Temperature is a measure of the average kinetic energy of the parti-
cles in a sample of matter. The greater the kinetic energy of the particles
in a sample, the higher the temperature is and the hotter it feels. To
assign a numerical value to temperature, it is necessary to define a tem-
perature scale. For calculations in thermochemistry, we use the Celsius
and Kelvin scales. Celsius and Kelvin temperatures are related by the
following equation.

K = 273.15 + °C

For most calculations in this book, 273.15 is rounded to 273.
The ability to measure temperature is thus based on energy transfer.
The amount of energy transferred as heat is usually measured in joules.
A joule is the SI unit of heat as well as all other forms of energy. The
joule, abbreviated J, is derived from the units for force and length.

kg × m2
N × m = s2

FIGURE 1 The direction Heat can be thought of as the energy transferred between samples of
of energy transfer is determined by matter because of a difference in their temperatures. Energy transferred as
the temperature differences between heat always moves spontaneously from matter at a higher temperature to
the objects within a system. The matter at a lower temperature, as shown in Figure 1. The temperature of
energy is transferred as heat from the cool water in the beaker increases as energy flows into it. Likewise,
the hotter brass bar to the cooler the temperature of the hot brass bar decreases as energy flows away from
water. This energy transfer will it.When the temperature of the water equals the temperature of the brass
continue until the bar and the water bar, energy is no longer transferred as heat within the system.
reach the same temperature.

Specific Heat

The quantity of energy transferred as heat during a temperature change
depends on the nature of the material changing temperature, the mass
of the material changing temperature, and the size of the temperature
change. One gram of iron heated to 100.0°C and cooled to 50.0°C in a
calorimeter transfers 22.5 J of energy to the surrounding water. But one
gram of silver transfers 11.8 J of energy under the same conditions.
The difference depends on the metals’ differing capacities for absorb-
ing this energy. A quantity called specific heat can be used to compare
heat absorption capacities for different materials. Specific heat is the
amount of energy required to raise the temperature of one gram of a
substance by one Celsius degree (1°C) or one kelvin (1 K)(because the
sizes of the degree divisions on both scales are equal). Values of spe-
cific heat can be given in units of joules per gram per Celsius degree,
J/(g•°C), joules per gram per kelvin, J/(g•K), or calories per gram per
Celsius degree, cal/(g•°C). Table 1 gives the specific heats of some
common substances. Notice the extremely high specific heat of water,
one of the highest of most common substances.

Specific heat is usually measured under constant pressure conditions,
so its symbol, cp, contains a subscripted p as a reminder to the reader.

532 C H A P T E R 1 6

TABLE 1 Specific Heats of Some Common
Substances at 298.15 K

Substance Specific heat J/(g•K)

Water (l) 4.18

Water (s) 2.06

Water (g) 1.87

Ammonia (g) 2.09

Benzene (l) 1.74

Ethanol (l) 2.44

Ethanol (g) 1.42

Aluminum (s) 0.897

Calcium (s) 0.647

Carbon, graphite (s) 0.709

Copper (s) 0.385

Gold (s) 0.129

Iron (s) 0.449

Mercury (l) 0.140

Lead (s) 0.129

In the following mathematical equation, cp is the specific heat at a given
pressure, q is the energy lost or gained, m is the mass of the sample, and
ΔT represents the difference between the initial and final temperatures.

q
cp = m × ΔT

This equation can be rearranged to give an equation that can be used to
find the quantity of energy gained or lost with a change in temperature.

q = cp × m × ΔT

SAMPLE PROBLEM A

A 4.0 g sample of glass was heated from 274 K to 314 K, a temperature increase of 40. K, and was found
to have absorbed 32 J of energy as heat.
a. What is the specific heat of this type of glass?
b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K?

SOLUTION

1 ANALYZE Given: m = 4.0 g
ΔT = 40. K
q = 32 J

Unknown: cp in J/(g•K)

R E A C T I O N E N E R G Y 533

2 PLAN a. The specific heat, cp, of the glass is calculated using the equation given for specific heat.
q

cp = m × ΔT
b. The rearranged specific heat equation is used to find the energy gained when the glass

was heated.

q = cp × m × ΔT

3 COMPUTE 32 J
a. = 0.20 J/(g•K)

(4.0 g)(40. K)

0.20 J
b. (4.0 g)(344 K − 314 K)

(g•K)

4 EVALUATE 0.20 J
(4.0 g)(30 K) = 24 J

(g•K)
The units combine or cancel correctly to give the specific heat in J/(g•K) and the energy in J.

PRACTICE Answers in Appendix E

1. Determine the specific heat of a material if a 35 g sample Go to go.hrw.com for
absorbed 96 J as it was heated from 293 K to 313 K. more practice problems
that ask you to calculate
2. If 980 kJ of energy are added to 6.2 L of water at 291 K, what will using specific heat.
the final temperature of the water be?
Keyword: HC6NRGX

Enthalpy of Reaction

The energy absorbed as heat during a chemical reaction at constant
pressure is represented by ΔH. The H is the symbol for a quantity called
enthalpy. It is not practical to talk just about enthalpy as a quantity,
because we have no way to directly measure the enthalpy of a system.
Only changes in enthalpy can be measured. The Greek letter Δ (“delta”)
stands for “change in.” Therefore, ΔH is read as “change in enthalpy.”
An enthalpy change is the amount of energy absorbed by a system as
heat during a process at constant pressure. The enthalpy change is always
the difference between the enthalpies of the products and the reactants.
The following equation expresses an enthalpy change for a reaction.

ΔH = Hproducts − Hreactants

The enthalpy of reaction is the quantity of energy transferred as heat dur-
ing a chemical reaction. You can think of enthalpy of reaction as the dif-
ference between the stored energy of the reactants and the products.
Enthalpy of reaction is sometimes called “heat of reaction.”

534 C H A P T E R 1 6

If a mixture of hydrogen and oxygen is ignited, water will form and www.scilinks.org
energy will be released explosively. The energy that is released comes Topic: Thermochemistry
from the reactants as they form products. Because energy is released, Code: HC61513
the reaction is exothermic, and the energy of the product, water, must be
less than the energy of the reactants. The following chemical equation R E A C T I O N E N E R G Y 535
for this reaction indicates that when 2 mol of hydrogen gas at room tem-
perature are burned, 1 mol of oxygen gas is consumed and 2 mol of
water vapor are formed.

2H2(g) + O2(g) ⎯→ 2H2O(g)

The equation does not tell you that energy is evolved as heat during the
reaction. Experiments have shown that 483.6 kJ of energy are evolved
when 2 mol of gaseous water are formed from its elements at 298.15 K.

Modifying the chemical equation to show the amount of energy as
heat released during the reaction gives the following expression.

2H2(g) + O2(g) ⎯→ 2H2O(g) + 483.6 kJ

This expression is an example of a thermochemical equation, an
equation that includes the quantity of energy released or absorbed as heat
during the reaction as written. In any thermochemical equation, we must
always interpret the coefficients as numbers of moles and never as
numbers of molecules. The quantity of energy released as heat in this or
any other reaction depends on the amounts of reactants and products.
The quantity of energy as heat released during the formation of water
from H2 and O2 is proportional to the quantity of water formed.
Producing twice as much water vapor would require twice as many
moles of reactants and would release 2 × 483.6 kJ of energy as heat, as
shown in the following thermochemical equation (which is simply the
previous thermochemical equation, multiplied by two).

4H2(g) + 2O2(g) ⎯→ 4H2O(g) + 967.2 kJ

Producing one-half as much water would require one-half as many moles
1⎯⎯
of reactants and would release only one-half as much energy, or 2 × 483.6 kJ.

The thermochemical equation for this reaction would be as follows.

H2(g) + 1 O2(g) ⎯→ H2O(g) + 241.8 kJ
2

The situation is reversed in an endothermic reaction because products
have a larger enthalpy than reactants. The decomposition of water vapor
is endothermic; it is the reverse of the reaction that forms water vapor.The
amount of energy as heat absorbed by water molecules to form hydrogen
and oxygen equals the amount of energy as heat released when the ele-
ments combine to form the water. This is to be expected because the dif-
ference between the energy of reactants and products is unchanged.
Enthalpy now appears on the reactant side of the thermochemical equa-
tion that follows, indicating that it is an endothermic reaction.

FIGURE 2 In an exothermic Exothermic Reaction Pathway Initial enthalpy
chemical reaction, the enthalpy
change is negative, meaning energy Reactants
is released from the system as heat.

Enthalpy Energy as ΔH
heat is evolved
ΔH is
negative

Final enthalpy Products
Course of reaction

www.scilinks.org 2H2O(g) + 483.6 kJ ⎯→ 2H2(g) + O2(g)
Topic: Enthalpy
Code: HC60522 The physical states of reactants and products must always be included
in thermochemical equations because they influence the overall
536 C H A P T E R 1 6 amount of energy as heat gained or lost. For example, the energy
needed for the decomposition of water would be greater than 483.6 kJ
if we started with ice, because extra energy would be needed to melt the
ice and to change the liquid into a vapor.

Thermochemical equations are usually written by designating the
value of ΔH rather than writing the energy as a reactant or product.
For an exothermic reaction, ΔH is always negative because the system
loses energy. So, the thermochemical equation for the exothermic for-
mation of 2 mol of gaseous water from its elements now has the fol-
lowing form.

2H2(g) + O2(g) ⎯→ 2H2O(g) ΔH = –483.6 kJ

Figure 2 graphically shows the course of an exothermic reaction. The
initial enthalpy of the reactants is greater than the final enthalpy of the
products. This means energy as heat is evolved, or given off, during the
reaction; this is described as a negative enthalpy change.

For an endothermic reaction, ΔH is always positive because the sys-
tem gains energy. Thus, the endothermic decomposition of 2 mol of
gaseous water has the following thermochemical equation.

2H2O(g) ⎯→ 2H2(g) + O2(g) ΔH = +483.6 kJ

The course of an endothermic reaction is illustrated in Figure 3. Energy
as heat is absorbed in this reaction, meaning that the initial enthalpy of

Endothermic Reaction Pathway Products FIGURE 3 In an endothermic
chemical reaction, the enthalpy
Final enthalpy ΔH is change is positive because energy is
positive absorbed into the system as heat.
Energy as
Enthalpy heat is absorbed ΔH

Reactants Initial enthalpy
Course of reaction

the reactants is lower than the final enthalpy of the products. In this
case, ΔH is designated as positive.

Keep in mind the following when using thermochemical equations.

1. The coefficients in a balanced thermochemical equation represent
the numbers of moles of reactants and products and never the num-
bers of molecules. This allows us to write these coefficients as frac-
tions rather than whole numbers when necessary.

2. The physical state of the product or reactant involved in a reaction
is an important factor and therefore must be included in the ther-
mochemical equation.

3. The change in enthalpy represented by a thermochemical equation is
directly proportional to the number of moles of substances undergo-
ing a change. For example, if 2 mol of water are decomposed, twice
as much enthalpy, 483.6 kJ, is needed than for the decomposition of
1 mol of water.

4. The value of the enthalpy change, ΔH, is usually not significantly
influenced by changing temperature.

Enthalpy of Formation

The formation of water from hydrogen and oxygen is a composition reac-
tion—the formation of a compound from its elements in their standard
form. Thermochemical data are often recorded as the enthalpies of such
composition reactions. The molar enthalpy of formation is the enthalpy
change that occurs when one mole of a compound is formed from its
elements in their standard state at 25ºC and 1 atm.

R E A C T I O N E N E R G Y 537

To make comparisons meaningful, enthalpies of formation are given

for the standard states of reactants and products—these are the states

found at atmospheric pressure and, usually, room temperature (298.15 K).

Thus, the standard state of water is liquid, not gas or solid. The standard

state of iron is solid, not a molten liquid.To signify that a value represents
measurements on substances in their standard states, a 0 sign is added to
the enthalpy symbol, giving ΔH0 for the standard enthalpy of a reaction.
Adding a subscript f, as in ΔHf0, further indicates a standard enthalpy of
formation.

Some standard enthalpies of formation are given in Appendix

Table A-14. Each entry in the table is the enthalpy of formation for the

synthesis of one mole of the compound listed from its elements in their

standard states. The thermochemical equation to accompany each

enthalpy of formation shows the formation of one mole of the compound

from its elements in their standard states.

Stability and Enthalpy of Formation

If a large amount of energy as heat is released when a compound is

formed, the compound has a large negative enthalpy of formation. Such

compounds are very stable.

Elements in their standard states are defined as having ΔH 0 = 0. The
f
0
ΔH f of carbon dioxide is −393.5 kJ/mol of gas produced. Therefore, car-

bon dioxide is more stable than the elements from which it was formed.

You can see in Appendix Table A-14 that the majority of the enthalpies

of formation are negative.

Compounds with relatively positive values of enthalpies of forma-

tion, or only slightly negative values, are typically unstable. For example,

hydrogen iodide, HI, is a colorless gas that decomposes somewhat when

stored at room temperature. It has a relatively high positive enthalpy of

formation of +26.5 kJ/mol. As it decomposes, violet iodine vapor, I2,
becomes visible throughout the container of the gas.

Compounds with a high positive enthalpy of formation are sometimes

very unstable and may react or decompose violently. For example, ethyne

(acetylene), C2H2 (ΔH 0 = +226.7 kJ/mol), reacts violently with oxygen
f

and must be stored in cylinders as a solution in acetone. Mercury fulmi-

nate, HgC2N2O2, has a very large enthalpy of formation of +270 kJ/mol.
Its instability makes it useful as a detonator for explosives.

Enthalpy of Combustion

Combustion reactions produce a considerable amount of energy in the
form of light and heat when a substance is combined with oxygen. The
enthalpy change that occurs during the complete combustion of one
mole of a substance is call the enthalpy of combustion of the substance.

538 C H A P T E R 1 6

Thermometer Stirrer Ignition wires FIGURE 4 This is a diagram of a
Water Steel bomb simple combustion calorimeter. A
Insulated Sample dish weighed sample is ignited by
outside an electric spark and burned in the
container sample dish in an atmosphere of
pure oxygen. The energy generated
Steel by the combustion reaction warms
container the steel bomb and the water
surrounding it. The thermometer
measures the initial and final
temperatures of the water, and this
temperature change is then used to
calculate the energy evolved by the
reaction as heat.

Enthalpy of combustion is defined in terms of one mole of reactant,
whereas the enthalpy of formation is defined in terms of one mole of
product. All substances are in their standard states. The general
enthalpy notation, ΔH, applies to enthalpies of reaction, but the addi-
tion of a subscripted c, ΔHc, refers specifically to enthalpy of combus-
tion. A list of enthalpies of combustion can be found in Appendix
Table A-5. A combustion calorimeter is a common instrument used to
determine enthalpies of combustion. Figure 4 shows a fixed-volume
calorimeter. A similar apparatus under constant pressure is used to
obtain enthalpy measurements.

Calculating Enthalpies of Reaction

Thermochemical equations can be rearranged and added to give enthalpy
changes for reactions not included in the data tables. The basis for cal-
culating enthalpies of reaction is known as Hess’s law: The overall
enthalpy change in a reaction is equal to the sum of enthalpy changes for
the individual steps in the process. The energy difference between reac-
tants and products is independent of the route taken to get from one to
the other. In fact, measured enthalpies of reaction can be combined to
calculate enthalpies of reaction that are difficult or impossible to actu-
ally measure.

To demonstrate how to apply Hess’s law, we will work through the
calculation of the enthalpy of formation for the formation of methane
gas, CH4, from its elements, hydrogen gas and solid carbon (graphite),
at 298.15 K (25°C).

C(s) + 2H2(g) ⎯→ CH4(g) ΔH 0 = ?
f

R E A C T I O N E N E R G Y 539

In order to calculate the change in enthalpy for this reaction, we can use
the combustion reactions of the elements, carbon and hydrogen, and of
methane.

C(s) + O2(g) ⎯→ CO2(g) ΔH 0 = −393.5 kJ
c

H2(g) + 1 O2(g) ⎯→ H2O(l ) ΔH 0 = −285.8 kJ
2 c

CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O(l) ΔH 0 = −890.8 kJ
c

The general principles for combining thermochemical equations follow.

1. If a reaction is reversed, the sign of ΔH is also reversed.
2. Multiply the coefficients of the known equations so that when added

together they give the desired thermochemical equation. Multiply
the ΔH by the same factor as the corresponding equation.

In this case, reverse the combustion equation for methane, and remem-
ber to change the sign of ΔH from negative to positive. This will change
the exothermic reaction to an endothermic one.

CO2(g) + 2H2O(l) ⎯→ CH4(g) + 2O2(g) ΔH0 = +890.8 kJ

Now we notice that 2 moles of water are used as a reactant; therefore,
2 moles of water will be needed as a product. In the combustion reac-
tion for hydrogen as it is written, it only produces one mole of water. We
must multiply the coefficients of this combustion reaction and the value
of ΔH by 2 in order to obtain the desired quantity of water.

2H2(g) + O2(g) ⎯→ 2H2O(l) ΔH 0 = 2(−285.8 kJ)
c

We are now ready to add the three equations together using Hess’s law
to give the enthalpy of formation for methane and the balanced equation.

C(s) + O2(g) ⎯→ CO2(g) ΔH 0 = −393.5 kJ
2H2(g) + O2(g) ⎯→ 2H2O(l) c
CO2(g) + 2H2O(l) ⎯→ CH4(g) + 2O2(g)
C(s) + 2H2(g) ⎯→ CH4(g) ΔH 0 = 2(−285.8 kJ)
c

ΔH0 = +890.8 kJ

ΔH 0 = −74.3 kJ
f

Hess’s law says that the enthalpy difference between reactants and
products is independent of pathway. Therefore, any enthalpy of reaction
may be calculated using enthalpies of formation for all the substances
in the reaction of interest, without knowing anything else about how the
reaction occurs. Mathematically, the overall equation for enthalpy
change will be in the form of the equation shown below.

ΔH0 = sum of [(ΔH 0 of products) × (mol of products)] −
f
0
sum of [(ΔH f of reactants) × (mol of reactants)]

An example using Hess’s law is shown in Sample Problem B.

540 C H A P T E R 1 6

SAMPLE PROBLEM B For more help, go to the Math Tutor at the end of this chapter.

Calculate the enthalpy of reaction for the combustion of nitrogen monoxide gas, NO, to form nitrogen
dioxide gas, NO2, as given in the following thermochemical equation.

NO(g) + 1 O2(g) ⎯→ NO2(g)
2

Use the enthalpy-of-formation data in Appendix Table A-14. Solve by combining the known thermochem-
ical equations. Verify the result by using the general equation for finding enthalpies of reaction from
enthalpies of formation.

SOLUTION

1 ANALYZE Given: 1 N2(g) + 1 O2(g) ⎯→ NO(g) ΔH 0 = +90.29 kJ
2 2 f

1 N2(g) + O2(g) ⎯→ NO2(g) ΔH 0 = +33.2 kJ
2 f

Unknown: ΔH 0 for NO(g) + 1 O2(g) ⎯→ NO2(g)
2

2 PLAN ΔH can be found by adding the ΔHs of the component reactions as specified in Hess’s law.

The desired equation has NO(g) and 1 O2(g) as reactants and NO2(g) as the product.
2

NO(g) + 1 O2(g) ⎯→ NO2(g)
2

We need an equation with NO as a reactant. Reversing the first reaction for the formation
of NO from its elements and the sign of ΔH yields the following thermochemical equation.

NO(g) ⎯→ 1 N2(g) + 1 O2(g) ΔH0 = −90.29 kJ
2 2

The other equation should have NO2 as a product, so we can retain the second equation for
the formation of NO2 from its elements as it stands.

1 N2(g) + O2(g) ⎯→ NO2(g) ΔH 0 = +33.2 kJ
2 f

NO(g) ⎯→ 1 N2(g) + 1 O2(g) ΔH0 = −90.29 kJ
2 2
3 COMPUTE
4 EVALUATE 1 N2(g) + O2(g) ⎯→ NO2(g) ΔH 0 = +33.2 kJ
2 f

NO(g) + 1 O2(g) ⎯→ NO2(g) ΔH0 = −57.1 kJ
2

Note the cancellation of the 1 N2(g) and the partial cancellation of the O2(g).
2

The unnecessary reactants and products cancel to give the desired equation.
The general relationship between the enthalpy of a reaction and the enthalpies of forma-
tion of the reactants and products is described in the following word equation.

ΔH 0 = sum of [(ΔH 0 of products) × (mol of products)] −
f
0
sum of [(ΔH f of reactants) × (mol of reactants)]

R E A C T I O N E N E R G Y 541

To find the necessary sums, the ΔH 0 value for each reactant and each product must be mul-
f

tiplied by its respective coefficient in the desired equation. For the reaction of NO with O2,
applying this equation gives the following value for ΔH0.

ΔH 0 = ΔH 0 (NO2) − [ΔH 0 (NO) + 0]
f f

= +33.2 kJ/mol − 90.29 kJ/mol = −57.1 kJ

Note that zero is the assigned value for the enthalpies of formation of elements in their
standard states.

PRACTICE Answers in Appendix E

1. Calculate the enthalpy of reaction for the combustion of methane
gas, CH4, to form CO2(g) + H2O(l).

2. Carbon occurs in two distinct forms. It can be the soft, black mate-
rial found in pencils and lock lubricants, called graphite, or it can
be the hard, brilliant gem we know as diamond. Calculate ΔH0 for
the conversion of graphite to diamond for the following reaction.

Cgraphite(s) ⎯→ Cdiamond(s)

The combustion reactions you will need follow. Go to go.hrw.com for
more practice problems
Cgraphite(s) + O2(g) ⎯→ CO2(g) ΔH 0 = −394 kJ that ask you to use
Cdiamond(s) + O2(g) ⎯→ CO2(g) c Hess’s law.
0
ΔH c = −396 kJ Keyword: HC6NRGX

Determining Enthalpy of Formation

When carbon is burned in a limited supply of oxygen, carbon monoxide
is produced. In this reaction, carbon is probably first oxidized to carbon
dioxide. Then part of the carbon dioxide is reduced with carbon to give
some carbon monoxide. Because these two reactions occur simultane-
ously and we get a mixture of CO and CO2, it is not possible to directly
measure the enthalpy of formation of CO(g) from C(s) and O2(g).

C(s) + 1 O2(g) ⎯→ CO(g) ΔH 0 = ?
2 f

However, we do know the enthalpy of formation of carbon dioxide and
the enthalpy of combustion of carbon monoxide.

C(s) + O2(g) ⎯→ CO2(g) ΔH 0 = −393.5 kJ/mol
f

CO(g) + 1 O2(g) ⎯→ CO2(g) ΔH 0 = −283.0 kJ/mol
2 c

We reverse the second equation because we need CO as a product.
Adding gives the desired enthalpy of formation of carbon monoxide.

542 C H A P T E R 1 6

Enthalpies of Reaction 0 kJ FIGURE 5 This diagram shows the
ΔH = −110.5 kJ enthalpy of reaction for carbon diox-
C(s) + O2(g) ide, CO2, and carbon monoxide, CO.
ΔH = −283.0 kJ
CO(g) + O2(g)
−393.5 kJ
ΔH = +283.0 kJ
CO2(g)

C(s) + O2(g) ⎯→ CO2(g) ΔH0 = −393.5 kJ
ΔH0 = +283.0 kJ
CO2(g) ⎯→ CO(g) + 1 O2(g) ΔH0 = −110.5 kJ
2

C(s) + 1 O2(g) ⎯→ CO(g)
2

Figure 5 is a model for the process described in this section. If we plot

the reactions based on their relative energy, you can see the relationship

among the values obtained for the enthalpy of formation of carbon

monoxide. The formation of CO2 is plotted at a level corresponding to
−393.5 kJ/mol. The diagram shows the reverse of the combustion reaction
(+283.0 kJ/mol) is added to that level. From the diagram, you see the dif-
ference, which represents the formation of CO.This value is −110.5 kJ/mol.

SAMPLE PROBLEM C For more help, go to the Math Tutor at the end of this chapter.

Calculate the enthalpy of formation of pentane, C5H12, using the information on enthalpies of formation in
Appendix Table A-14 and the information on enthalpies of combustion in Appendix Table A-5. Solve by

combining the known thermochemical equations.

SOLUTION

1 ANALYZE Given: C(s) + O2(g) ⎯→ CO2(g) ΔH 0 = −393.5 kJ
f

H2(g) + 1 O2(g) ⎯→ H2O(l ) ΔH 0 = −285.8 kJ
2 f

C5H12(g) + 8O2(g) ⎯→ 5CO2(g) + 6H2O(l) ΔH 0 = −3535.6 kJ
c

Unknown: ΔH 0 for 5C(s) + 6H2(g) ⎯→ C5H12(g)
f

R E A C T I O N E N E R G Y 543

2 PLAN Combine the given equations according to Hess’s law. We need C5H12 as a product, so we
reverse the equation for combustion of C5H12 and the sign for ΔHc0. Multiply the equation
for formation of CO2 by 5 to give 5C as a reactant. Multiply the equation for formation of
H2O by 6 to give 6H2 as a reactant.

3 COMPUTE 5C(s) + 5O2(g) ⎯→ 5CO2(g) ΔH0 = 5(−393.5 kJ)
6H2(g) + 3O2(g) ⎯→ 6H2O(l) ΔH0 = 6(−285.8 kJ)

5CO2(g) + 6H2O(l) ⎯→ C5H12(g) + 8O2(g) ΔH0 = +3535.6 kJ
5C(s) + 6H2(g) ⎯→ C5H12(g)
ΔH 0 = −146.7 kJ
f

4 EVALUATE The unnecessary reactants and products cancel to give the correct equation.

PRACTICE Answers in Appendix E

1. Calculate the enthalpy of formation of butane, C4H10, using the bal-
anced chemical equation and information in Appendix Table A-5 and

Table A-14. Write out the solution according to Hess’s law.

2. Calculate the enthalpy of combustion of 1 mol of nitrogen, N2, to
form NO2 using the balanced chemical equation and Appendix
Table A-14.

3. Calculate the enthalpy of formation for sulfur dioxide, SO2, from Go to go.hrw.com for
its elements, sulfur and oxygen. Use the balanced chemical equa- more problems that ask
you to calculate
tion and the following information. enthalpies of formation
and combustion.
S(s) + 3 O2(g) ⎯→ SO3(g) ΔH 0 = −395.2 kJ
2 c Keyword: HC6NRGX

2SO2(g) + O2(g) ⎯→ 2SO3(g) ΔH0 = −198.2 kJ

SECTION REVIEW Critical Thinking

1. What is meant by enthalpy change? 6. INTEGRATING CONCEPTS Isooctane (C8H18) is a
major component of gasoline.
2. What is meant by enthalpy of reaction?
a. Using the following thermodynamic data, calcu-
3. Describe the relationship between a compound’s
stability and its enthalpy of formation. late the change in enthalpy for the combustion

4. What is the importance of Hess’s law to thermo- of 1.0 mol of isooctane.
dynamic calculations?
H2(g )++O21⎯2⎯O(g2()g⎯) ⎯→→COH22(Og()gΔ)HΔ0H=0 = −241.8 kJ;
5. How much energy would be absorbed as heat by C(s ) −393.5 kJ;
75 g of iron when heated from 295 K to 301 K?
8C(s) + 9H2(g) ⎯→ C8H18(l ) ΔH0 = −224.13 kJ.

b. One gallon of isooctane has a mass of 2.6 kg.

What is the change in enthalpy for the combus-

tion of one gallon of this compound?

544 C H A P T E R 1 6

Self-Heating Meals

Who would have thought that This product uses supercorrosion to “We’ve made about 80 million
corrosion could be useful? The give you a hot meal. heaters for the military in the last 10
HeaterMeals Company did. years. Lately we’ve been successfully
This company uses the prop- 1930s. “But,” says McLandrich, “there marketing them to long-distance
erties of saltwater corrosion really have been no significant uses truck drivers. The product is in about
to heat TV-type dinners, and of the product until the Desert Storm 800 truck stops in 48 states.”
now it is taking packaged conflict, which led to the military’s
foods to a new level of taking this technology and adopting The company has plans to develop
convenience. it for field use so that soldiers could other products using the
heat a meal-ready-to-eat.
HeaterMeals’ products, as controlled use of
their name implies, come with “supercorrosion.” “A beverage
a self-contained heat source. Each could be heated,” says
meal contains a package of food, a McLandrich, “and we do have
tray that holds a porous pouch con- prototypes for a baby-bottle
taining Mg and Fe alloy, and a 2 oz warmer. We’re also working
pouch filled with salt water. When on making a portable hot cup
the salt water is poured into the tray of coffee or a hot cup of tea
with the porous pouch, it or cocoa.”
begins to vigorously corrode
the metals. The sealed, pre- Questions
cooked food package is then
placed on top of the tray and 1. How did the development of
returned to its box, where the self-heating metallic alloy pow-
temperature of the food pack- ders benefit the military?
age is raised by 100°F, heating
the meal in 14 min. 2. If the temperature of 50.0 mL
of water is raised from 25.0ºC to
Corrosion, the process by 100.ºC using the self-heating
which a metal reacts with air meal package, what is the change
or water, is usually an undesir- in the enthalpy of the magnesium
able event, such as when iron cor- reaction in this package?
rodes to form rust. With
HeaterMeals, however, the corrosion www.scilinks.org
process is speeded up to produce an Topic: Supercorrosion
exothermic reaction—with the Code: HC61479
excess energy as the desired result.

According to Drew McLandrich, of
the HeaterMeals Company, the idea
for using self-heating metallic alloy
powders has been around since the

R E A C T I O N E N E R G Y 545

SECTION 2 Driving Force
of Reactions
OBJECTIVES
T he change in energy of a reaction system is one of two factors that
Explain the relationship
between enthalpy change allow chemists to predict whether a reaction will occur spontaneously
and the tendency of a and to explain how it occurs. The randomness of the particles in a
reaction to occur. system is the second factor affecting whether a reaction will occur spon-
taneously.
Explain the relationship
between entropy change Enthalpy and Reaction Tendency
and the tendency of a
reaction to occur. The great majority of chemical reactions in nature are exothermic. As
these reactions proceed, energy is liberated and the products have less
Discuss the concept of free energy than the original reactants. The products are also more resistant
energy, and explain how the to change, more stable, than the original reactants. The tendency
value of this quantity is calcu- throughout nature is for a reaction to proceed in a direction that leads
lated and interpreted. to a lower energy state.

Describe the use of free We might think that endothermic reactions, in which energy is
energy change to determine absorbed, cannot occur spontaneously because the products are at
the tendency of a reaction higher potential energy and are less stable than the original reactants.
to occur. They would be expected to proceed only with the assistance of an out-
side influence, such as continued heating. However, some endothermic
reactions do occur spontaneously. We conclude that something other
than enthalpy change must help determine whether a reaction will occur.

www.scilinks.org Entropy and Reaction Tendency
Topic: Entropy
Code: HC60523 A naturally occurring endothermic process is melting. An ice cube melts
spontaneously at room temperature as energy is transferred from the
546 C H A P T E R 1 6 warm air to the ice. The well-ordered arrangement of water molecules
in the ice crystal is lost, and the less-ordered liquid phase of higher ener-
gy content is formed. A system that can go from one state to another
without a decrease in enthalpy does so with an increase in entropy.

Look at the physical states of the reactants in the chemical equation
for the decomposition of ammonium nitrate.

2NH4NO3(s) ⎯→ 2N2(g) + 4H2O(l) + O2(g)

On the left side are 2 mol of solid ammonium nitrate. The

right-hand side of the equation shows 3 mol of gaseous

molecules plus 4 mol of a liquid. The arrangement of par-

ticles on the right-hand side of the equation is more ran-

dom than the arrangement on the left side of the

equation and hence is less ordered. Figures 6a and 6b

show the reactant and products of this decomposition

reaction.

These examples illustrate that there is a tendency in

nature to proceed in a direction that increases the ran-

domness of a system. A random system is one that lacks a

regular arrangement of its parts. This tendency toward

randomness is called entropy. Entropy, S, can be defined

in a simple qualitative way as a measure of the degree of

randomness of the particles, such as molecules, in a system.

To understand the concept of entropy, consider solids, liq-

uids, and gases. In a solid, the particles are fixed in posi-

tion in their small regions of space, but they are vibrating (a)

back and forth. Even so, we can determine with fair pre-

cision the location of the particles. The degree of ran-

domness is low, so the entropy is low. When the solid

melts, the particles are still very close together, but they

can move about somewhat. The system is more random,

and it is more difficult to describe the location of the par-

ticles. The entropy is higher. When the liquid evaporates,

the particles are moving rapidly and are also much farther

apart. Locating an individual particle is much more diffi-

cult, and the system is much more random. The entropy of

the gas is still higher than that of the liquid. A general but

not absolute rule is that the entropy of liquids is larger

than that of solids and the entropy of gases is larger than

that of liquids. But this rule must be used with caution.

For example, the entropy of liquid mercury is much less

than that of some solids.

The entropy of a pure crystalline solid is zero at

absolute zero. As energy is added, the randomness of the (b)
molecular motion increases. Measurements of energy
absorbed and calculations are used to determine the FIGURE 6 When ammonium nitrate,
absolute entropy or standard molar entropy, and values NH4NO3, decomposes, the entropy of the reaction
are then recorded in tables. These molar values are system increases as (a) one solid reactant becomes
reported as kJ/(mol•K). Entropy change, which can also (b) two gaseous products and one liquid product.

be measured, is defined as the difference between the

entropy of the products and the reactants. Therefore, an increase in

entropy is represented by a positive value for ΔS, and a decrease in

entropy is represented by a negative value for ΔS.

The process of forming a solution almost always involves an increase

in entropy because there is an increase in randomness. This is true for

mixing gases, dissolving a liquid in another liquid, and dissolving a solid

in a liquid.

R E A C T I O N E N E R G Y 547

Water molecule Figure 7 illustrates the entropy change that takes

place when solid sugar is dissolved in tea (an aqueous

solution). In the sugar-water system shown in

Figure 7a, the solid sugar has just been added to the

Sugar tea, but most of it has not yet dissolved. The entropy
molecules is low because the majority of the sugar molecules are

in one region at the bottom of the pitcher and the

majority of the water molecules can be found every-

Low entropy where else in the pitcher. After the sugar dissolves in

(a) the tea, shown in Figure 7b, the sugar molecules are

Sugar molecules thoroughly mixed throughout the tea solution. Sugar
molecules and water molecules might be found any-

where in the solution, so the entropy, the randomness,

of the system increases. This would give ΔS a positive

value for this solid-liquid system. You can imagine the

Water same series of events happening for a system of gases
molecule mixing with each other or a system of liquids mixing.

In each case, ΔS would have a positive value once the

High entropy solution was formed.

(b)

FIGURE 7 When a solid Free Energy
dissolves in a liquid, the entropy
of the system increases. Processes in nature are driven in two directions: toward least enthalpy
and toward largest entropy. When these two oppose each other, the
CROSS-DISCIPLINARY dominant factor determines the direction of change. As a way to predict
which factor will dominate for a given system, a function has been
Go to go.hrw.com for a full-length defined to relate the enthalpy and entropy factors at a given tempera-
article on free-energy changes ture and constant pressure. This combined enthalpy-entropy function is
related to ATP in the body. called the free energy, G, of the system; it is also called Gibbs free ener-
gy. This function simultaneously assesses the tendencies for enthalpy
Keyword: HC6NRGX and entropy to change. Natural processes proceed in the direction that
lowers the free energy of a system.

Only the change in free energy can be measured. It can be defined in
terms of changes in enthalpy and entropy. At a constant pressure and
temperature, the free-energy change, ΔG, of a system is defined as the
difference between the change in enthalpy, ΔH, and the product of the
Kelvin temperature and the entropy change, which is defined as TΔS.

ΔG0 = ΔH 0 − TΔS0

Note that this expression is for substances in their standard states. The
product TΔS and the quantities ΔG and ΔH have the same units, usually
kJ. The units of ΔS for use in this equation are usually kJ/K. If ΔG < 0,
the reaction is spontaneous.

ΔH and ΔS in the free-energy equation can have positive or negative
values. This leads to four possible combinations of terms.

Table 2 on the next page shows that if ΔH is negative and ΔS is
positive, then both terms on the right in the free energy equation are

548 C H A P T E R 1 6

negative. Both factors contribute to the process’s being spontaneous. Diamonds Are Forever?
Therefore, ΔG will always be negative, and the reaction is definitely
spontaneous. On the other hand, if ΔH is positive (endothermic process) Carbon occurs in different forms, two of
and ΔS is negative (decrease in randomness), then the reaction as writ- which are graphite and diamond. Using
ten is not spontaneous. When the enthalpy and entropy changes are thermodynamic data, the change in
operating in different directions, sometimes one will predominate and free energy for diamond converting to
sometimes the other will predominate. There are reactions in which the graphite under standard thermodynam-
enthalpy change is negative and the entropy change is negative. The ic conditions is −3 kJ/mol. That is, since
enthalpy factor leads to a spontaneous process, but the negative entropy ΔG is negative for this reaction, dia-
change opposes this. This is true in the following reaction. The entropy mond should spontaneously change to
decreases because there is a decrease in moles of gas. graphite at 25ºC and 1 atm. So, why
doesn’t all of our diamond jewelry
C2H4(g) + H2(g) ⎯→ C2H6(g) change to graphite? The reaction rate is
too slow for this spontaneous change
There is a fairly large decrease in entropy, ΔS0 = −0.1207 kJ/(mol•K). to be observed. Therefore, at 25ºC and
However, the reaction is strongly exothermic, with a ΔH0 = −136.9 kJ/mol. 1 atm, although diamonds are not “for-
The reaction proceeds because the enthalpy term predominates. ever,” they will last a very long time.

ΔG0 = ΔH0 − TΔS0 = −136.9 kJ/mol − 298 K[−0.1207 kJ/(mol•K)]
= −100.9 kJ/mol

We can contrast this with the common commercial process for the
manufacture of syngas, a mixture of CO and H2. (This gas mixture is the
starting point for the synthesis of a number of large-volume commercial
chemicals, such as methanol, CH3OH.)

CH4(g) + H2O(g) ⎯→ CO(g) + 3H2(g)

This reaction is endothermic, with ΔH0 = +206.1 kJ/mol and ΔS0 =
+0.215 kJ/(mol•K), at standard conditions. The resulting ΔG is positive
at room temperature. This tells us that the reaction will not occur at
room temperature even though the entropy change is favorable.

ΔG0 = ΔH0 − TΔS0 = +206.1 kJ/mol − 298 K[+0.215 kJ/(mol•K)]
= +142.0 kJ/mol

TABLE 2 Relating Enthalpy, Entropy, and Free-
Energy Changes to Reaction Occurrence

⌬H ⌬S ⌬G
always negative
− value + value
(exothermic) (more random) negative at lower
temperatures
− value − value negative at higher
(exothermic) (less random) temperatures
never negative
+ value + value
(endothermic) (more random)

+ value − value
(endothermic) (less random)

R E A C T I O N E N E R G Y 549

SAMPLE PROBLEM D

For the reaction NH4Cl(s) ⎯→ NH3(g) + HCl(g), at 298.15 K, ⌬H0 = 176 kJ/mol and
⌬S0 = 0.285 kJ/(mol•K). Calculate ⌬G0, and tell whether this reaction is spontaneous in
the forward direction at 298.15 K.

SOLUTION

1 ANALYZE Given: ΔH0 = 176 kJ/mol at 298.15 K
ΔS0 = 0.285 kJ/(mol•K) at 298.15 K

Unknown: ΔG0 at 298.15 K

2 PLAN ΔS, ΔH, T → ΔG

The value of ΔG can be calculated according to the following equation.

3 COMPUTE ΔG0 = ΔH 0 − TΔS0

ΔG0 = 176 kJ/mol − 298 K [0.285 kJ/(mol•K)]
= 176 kJ/mol − 84.9 kJ/mol
= 91 kJ/mol

4 EVALUATE The answer is reasonably close to an estimated value of 110, calculated as 200 − (300 × 0.3).
The positive value of ΔG shows that this reaction does not occur naturally at 298.15 K.

PRACTICE Answers in Appendix E

1. For the vaporization reaction Br2(l) ⎯→ Br2(g), ΔH0 = 31.0 kJ/mol Go to go.hrw.com for
and ΔS0 = 93.0 J/(mol•K). At what temperature will this process be more practice problems
spontaneous? that ask you to
determine spontaneity
based on free-energy
changes.

Keyword: HC6NRGX

SECTION REVIEW 6. In the reaction in Sample Problem D, why does the
entropy increase?
1. What kind of enthalpy change favors a sponta-
neous reaction? Critical Thinking

2. What is entropy, and how does it relate to spon- 7. APPLYING MODELS Most biological enzymes
taneity of reactions? become denatured when they are heated and lose
their ability to catalyze reactions. This process
3. List several changes that result in an entropy (original enzyme ⎯→ denatured enzyme) is
increase. endothermic and spontaneous. Which structure,
the original enzyme or the denatured enzyme, is
4. Define free energy, and explain how its change more ordered? Explain your reasoning using ther-
is calculated. modynamic concepts.

5. Explain the relationship between free-energy
change and spontaneity of reactions.

550 C H A P T E R 1 6

CHAPTER HIGHLIGHTS

Thermochemistry • Thermochemistry is the study of the changes in energy that

Vocabulary accompany chemical reactions and physical changes.
thermochemistry
calorimeter • A thermochemical equation is an equation that includes the
temperature
joule quantity of energy released or absorbed as heat during the
heat
specific heat reaction as written.
enthalpy change
enthalpy of reaction • An enthalpy change is the amount of energy absorbed as heat
thermochemical equation
molar enthalpy of formation by a system in a process carried out at constant pressure.
enthalpy of combustion
Hess’s law • The enthalpy of reaction is the enthalpy change that occurs

during a chemical reaction.

• The enthalpy change is negative for exothermic reactions and

positive for endothermic reactions.

• Compounds with highly negative enthalpies of formation tend

to be stable; compounds with highly positive or only slightly

negative enthalpies of formation tend to be unstable.

• The standard molar enthalpy of formation is the enthalpy

change that occurs when one mole of a compound is formed
from its elements in their standard states at 25°C and 1 atm.

• The enthalpy change that occurs in a combustion reaction is

called the enthalpy of combustion.

• Enthalpies of reaction can be calculated by using enthalpies of

formation of reactants and products.

Driving Force of Reactions

Vocabulary • The tendency throughout nature is for a reaction to proceed
entropy
free energy in the direction that leads to a lower energy state.
free-energy change
• Entropy is a measure of the randomness of a system.
• Free-energy change combines the effects of entropy and

enthalpy changes and temperature of a system, and it is a mea-

sure of the overall tendency toward natural change.

• A reaction is spontaneous if it is accompanied by a decrease in

free energy. It is not spontaneous if there is an increase in free

energy.

R E A C T I O N E N E R G Y 551

CHAPTER REVIEW

For more practice, go to the Problem Bank in Appendix D. 12. Rewrite each equation below with the ΔH value

Thermochemistry included with either the reactants or the prod-

SECTION 1 REVIEW ucts, and identify the reaction as endothermic or

1. How does the enthalpy of the products of a exothermic.
reaction system compare with the enthalpy of
the reactants when the reaction is a. H2(g) + O2(g) ⎯→ H2O(l);
a. endothermic? ΔH0 = −285.83 kJ
b. exothermic?
b. 2Mg(s) + O2(g) ⎯→ 2MgO(s);
2. a. Distinguish between enthalpies of reaction, ΔH0 = −1200 kJ
formation, and combustion.
c. I2(s) ⎯→ I2(g); ΔH0 = +62.4 kJ
b. On what basis are enthalpies of formation d. 3CO(g) + Fe2O3(s) ⎯→ 2Fe(s) + 3CO2(g);
and combustion defined?
ΔH0 = −24.7 kJ
3. Write the equation that can be used to calculate 13. Use Appendix Table A-14 to write the reaction
the enthalpy of reaction from enthalpies of
formation. illustrating the formation of each of the follow-

4. What factors affect the value of ΔH in a reac- ing compounds from its elements. Write the ΔH
tion system? as part of each equation, and indicate the ΔH

5. State Hess’s law. How is it used? for the reverse reaction.
6. Describe a combustion calorimeter. What infor-
a. CaCl2(s)
mation can it give? b. C2H2(g) (ethyne, or acetylene)
c. SO2(g)
PRACTICE PROBLEMS 14. The reaction 2Fe2O3(s) + 3C(s) ⎯→
4Fe(s) + 3CO2(g) is involved in the smelting
7. How much energy is needed to raise the temper- of iron. Use ΔHf values given in Appendix
ature of a 55 g sample of aluminum from 22.4°C Table A-14 to calculate the enthalpy change
to 94.6°C? Refer to Table 1 for the specific heat
of aluminum. (Hint: See Sample Problem A.) during the production of 1 mol of iron.

8. If 3.5 kJ of energy are added to a 28.2 g sample 15. Use enthalpy-of-formation data given in
of iron at 20°C, what is the final temperature of
the iron in kelvins? Refer to Table 1 for the Appendix Table A-14 to calculate the enthalpy
specific heat of iron.
of reaction for each of the following. Solve each
9. You need 70.2 J to raise the temperature of 34.0 g
of ammonia, NH3(g), from 23.0°C to 24.0°C. by combining the known thermochemical equa-
Calculate the specific heat of ammonia.
tions. Verify each result by using the general
10. Calculate cp for indium metal, given that
1.0 mol In absorbs 53 J while increasing in equation for finding enthalpies of reaction from
temperature from 297.5 K to 299.5 K.
enthalpies of formation. (Hint: See Sample
11. For each equation listed below, determine the
ΔH and type of reaction (endothermic or Problem B.)
exothermic).
a. C(s) + O2(g) ⎯→ CO2(g) + 393.51 kJ a. CaCO3(s) ⎯→ CaO(s) + CO2(g)
b. CH4(g) + 2O2(g) ⎯→ b. Ca(OH)2(s) ⎯→ CaO(s) + H2O(g)
CO2(g) + 2H2O(l) + 890.31 kJ c. Fe2O3(s) + 3CO(g) ⎯→ 2Fe(s) + 3CO2(g)
c. CaCO3(s) + 176 kJ ⎯→ CaO(s) + CO2(g) 16. For glucose, C6H12O6(s), ΔHf = −1263 kJ/mol.
d. H2O(g) ⎯→ H2O(l) + 44.02 kJ Calculate the enthalpy change when 1 mol of

C6H12O6(s) combusts to form CO2(g) and
H2O(l ).
17. Calculate the standard enthalpies of reaction

for combustion reactions in which ethane, C2H6,
and benzene, C6H6, are the respective reactants
and CO2(g) and H2O(l) are the products in
each. Solve each by combining the known ther-

mochemical equations using the ΔHf values in

552 C H A P T E R 1 6

CHAPTER REVIEW

Appendix Table A-14. Verify the result by using reaction will occur spontaneously. (Hint: See
the general equation for finding enthalpies of
reaction from enthalpies of formation. Sample Problem D.)
a. C2H6(g) + O2(g) ⎯→
b. C6H6(l) + O2(g) ⎯→ a. ΔH = +125 kJ, T = 293 K,
18. The enthalpy of formation of ethanol, C2H5OH, ΔS = 0.0350 kJ/K
is −277.0 kJ/mol at 298.15 K. Calculate the
enthalpy of combustion of one mole of ethanol, b. ΔH = −85.2 kJ, T = 127°C,
assuming that the products are CO2(g) and ΔS = 0.125 kJ/K
H2O(l). (Hint: See Sample Problem C.)
c. ΔH = −275kJ, T = 773 K,
Driving Force of Reactions ΔS = 0.450 kJ/K

SECTION 2 REVIEW 28. The ΔS0 for the reaction shown, at 298.15 K, is
0.003 00 kJ/(mol•K). Calculate the ΔG0 for this
19. Would entropy increase or decrease for changes reaction, and determine whether it will occur
in state in which the reactant is a gas or liquid
and the product is a solid? What sign would the spontaneously at 298.15 K.
entropy change have?
C(s) + O2(g) ⎯→ CO2(g) + 393.51 kJ
20. How does an increase in temperature affect the
entropy of a system? MIXED REVIEW

21. What combination of ΔH and ΔS values always 29. When graphite reacts with hydrogen at 300 K, ΔH
produces a negative free-energy change?
is −74.8 kJ and ΔS is −0.0809 kJ/K. Will this reac-
22. Explain the relationship between temperature
and the tendency for reactions to occur tion occur spontaneously?
spontaneously.
30. How might you change reaction conditions to
PRACTICE PROBLEMS
induce an endothermic reaction that does not
23. A reaction has ΔH = −356 kJ and ΔS = −36 J/K.
Calculate ΔG at 25°C to confirm that the reac- occur naturally?
tion is spontaneous.
31. The diagram below represents an interpretation
24. A reaction has ΔH = 98 kJ and ΔS = 292 J/K.
Investigate the spontaneity of the reaction at of Hess’s law for the following reaction.
room temperature. Would increasing the tem-
perature have any effect on the spontaneity of Sn(s) + 2Cl2(g) ⎯→ SnCl4(l)
the reaction? Use the diagram to determine ΔH for each step

25. A reaction has ΔH = −76 kJ and ΔS = −117 J/K. and the net reaction.
Calculate ΔG for the reaction at 298.15 K. Is the
reaction spontaneous? Sn(s) + Cl2(g) ⎯→ SnCl2(s) ΔH = ?
SnCl2(l) + Cl2(g) ⎯→ SnCl4(s) ΔH = ?
26. The gas-phase reaction of H2 with CO2 to Sn(s) + 2Cl2(g) ⎯→ SnCl4(l) ΔH = ?
produce H2O and CO has ΔH = 11 kJ and
ΔS = 41 J/K. Is the reaction spontaneous at 0 Sn(s), Cl2(g)
298.15 K? What is ΔG? -100
-200 −325.1 kJ 50% phosphorus
27. Based on the following values, compute ΔG val- -300 fertilizer
ues for each reaction and predict whether the -400
-500 ?
-600
SnCl2(s) −186.2 kJ

SnCl4(l )

32. The standard enthalpy of formation for sulfur
dioxide gas is −296.8 kJ/mol. Calculate the
amount of energy given off in kJ when 30.0 g
of SO2(g) is formed from its elements.

R E A C T I O N E N E R G Y 553

CHAPTER REVIEW

33. The thermite reaction used in some welding 39. Interpreting Concepts Look at the two pic-
tures below. Which picture appears to have
applications has the following enthalpy and more order? Why? Are there any similarities
between the amount of order the marbles have
entropy changes at 298.15 K. Assuming ΔS and and the entropy of particles?
ΔH are constant, calculate ΔG at 448 K.
Fe2O3 + 2Al(s) ⎯→ 2Fe(s) + Al2O3(s) (a) (b)
ΔH0 = −851.5 kJ ΔS0 = −38.5 J/K
34. Rewrite each equation below with the ΔH value 40. Inferring Conclusions A reaction is endother-
included in either the reactants or products, mic and has a ΔH = 8 kJ. This reaction occurs
spontaneously at 25°C. What must be true about
and identify the reaction as endothermic or the entropy change?

exothermic. 41. Inferring Conclusions If both ΔH and ΔS
are negative, how does temperature affect
a. 2SO2(g) + O2(g) ⎯→ 2SO3(g); spontaneity?
ΔH = −197.8 kJ
42. Inferring Relationships If the reaction
b. 2NO2(g) ⎯→ 2NO(g) + O2(g); X ⎯→ Y is spontaneous, what can be said
ΔH = +114.2 kJ about the reaction Y ⎯→ X?

c. C2H4(g) + 3O2(g) ⎯→ 2CO2(g) + 2H2O(l); 43. Interpreting Concepts Absolute enthalpy can-
ΔH = −1411.0 kJ not be determined; only change in energy can
be measured. However, absolute entropy can be
35. Calculate the change in enthalpy for the follow- determined. Explain why an absolute entropy
can be determined.
ing reaction.

4FeO(s) + O2(g) ⎯→ 2Fe2O3(s)
Use the enthalpy-of-formation data listed in

Appendix Table A-14.

36. The reaction to synthesize methanol (CH3OH)
industrially is

CO(g) + 2H2(g) ⎯→ CH3OH(g).

The ΔH 0 = −90.7 kJ and the ΔS 0 =
reaction reaction
−220.8 J/K. At what temperatures is the reaction

nonspontaneous?

37. What is the main characteristic of a calorimeter

in a bomb calorimeter experiment, and why is

this characteristic essential?

CRITICAL THINKING

38. Relating Ideas Given the entropy change for
the first two reactions below, calculate the
entropy change for the third reaction below.
S8(s) + 8O2(s) ⎯→ 8SO2(g) ΔS = 89 J/K
2SO2(s) + O2(s) ⎯→ 2SO3(g) ΔS = −188 J/K
S8(s) + 12O2(s) ⎯→ 8SO3(g) ΔS = ?

554 C H A P T E R 1 6

CHAPTER REVIEW

RESEARCH & WRITING 46. Performance Design an experiment to mea-
sure the molar heat capacities of zinc and cop-
44. Obtain information on alternative units of mea- per. If your teacher approves the design, obtain
sure used to express values of energy as heat and the materials needed and conduct the experi-
other forms of energy. Also, find out how the ment. When you are finished, compare your
quantities relate to SI units. Include information experimental values with those from a chemical
specifically on English units, such as the British handbook or other reference source.
thermal unit (BTU), and on typical BTU ratings
of household appliances. Calculate how these rat- 47. Performance Develop a procedure to measure
ings would be expressed in joules instead. the ΔH of the reaction shown below. If your
teacher approves your procedure, test your pro-
ALTERNATIVE ASSESSMENT cedure by measuring the ΔH value of the reac-
tion. Determine the accuracy of your method by
45. Performance Design a simple calorimeter comparing your ΔH with the accepted ΔH value.
investigation to determine the molar enthalpy
of fusion of water. Use the following materials: CH3COONa(s) ⎯→ Na+(aq) + CH3COO−(aq)
a large plastic-foam cup with cover, a ther-
mometer, a balance, water at room temperature, Graphing Calculator
and an ice cube. Allow your teacher to review Free-Energy Changes
your design. Then carry out the investigation, Go to go.hrw.com for a graphing calculator
and write a laboratory report including your exercise that asks you to calculate and graph
calculations and a comparison of your quantita- free-energy (ΔG) changes.
tive results with known values. Try to account
for any disagreements between the experimen- Keyword: HC6NRGX
tal and actual values.

R E A C T I O N E N E R G Y 555

Math Tutor HESS’S LAW

You may have seen a popular comic strip in which a little boy takes a long, twisting path
between the school-bus stop and home. No matter which path the boy takes, the result
is always the same: He goes from the bus stop to the door of his house. Hess's law covers
a similar situation in thermochemistry. No matter which or how many steps occur in the
process of changing one or more substances into one or more other substances, the over-
all change in enthalpy is always the same. Hess's law can be used, for example, to predict
the enthalpy change, ΔH0, of a reaction without actually carrying out the reaction.

SAMPLE

Determine ΔH for the burning of carbon disulfide in oxygen.
CS2(l) + 3O2(g) ⎯→ CO2(g) + 2SO2(g) ΔH0 = ?

Use the following information:

C(s) + O2(g) ⎯→ CO2(g) ΔH 0 = −393.5 kJ/mol
f

S(s) + O2(g) ⎯→ SO2(g) ΔH 0 = −296.8 kJ/mol
f

C(s) + 2S(s) ⎯→ CS2(l ) ΔH 0 = 87.9 kJ/mol
f

Rearrange the given equations in a way that will put the reactants of the above equation

on the left and the products on the right.

1. C(s) + O2(g) ⎯→ CO2(g) ΔH 0 = −393.5 kJ/mol
f

2. 2S(s) + 2O2(g) ⎯→ 2SO2(g) ΔH0 = 2(−296.8 kJ)

3. CS2(l) ⎯→ C(s) + 2S(s) ΔH0 = −87.9 kJ

SUM: CS2(l) + 3O2(g) ⎯→ CO2(g) + 2SO2(g)

Notice that equation 2 is double the original equation S(s) + O2(g) ⎯→ SO2(g). The rea-
son for this is that 2SO2 are needed on the product side to match the 2SO2 in CS2(l) +
3O2(g) ⎯→ CO2(g) + 2SO2(g). The third equation is the reverse of the original, putting CS2
on the reactant side of the final equation. The sign of ΔH is likewise reversed. The value of
ΔH0 is the sum of the ΔH0 values for the three added equations.

ΔH0 = −393.5 kJ + 2(−296.8 kJ) + (−87.9 kJ)

ΔH0 = −1075.0 kJ

PRACTICE PROBLEMS

1. Calculate ΔH0 for the complete oxidation 2. Calculate ΔH0 for the reaction in which zinc
of sulfur to sulfur trioxide.
sulfide ore is roasted to obtain zinc oxide.

S(s) + 3 O2(g) ⎯→ SO3(g) ZnS(s) + 3 O2(g) ⎯→ ZnO(s) + SO2(g)
2 2

Use the following information. Use the following information.

S(s) + O2(g) ⎯→ SO2(g) ΔH 0 = −296.8 kJ/mol Zn(s) + 21O2(g) ⎯→ ZnO(s) ΔH 0 = −348.0 kJ/mol
f f

SO2(g) + 1 O2(g) ⎯→ SO3(g) ΔH 0 = −99.2 kJ/mol Zn(s) + S(s) ⎯→ ZnS(s) ΔH 0 = −203.0 kJ/mol
2 f f

S(s) + O2(g) ⎯→ SO2(g) ΔH 0 = −296.8 kJ/mol
f

556 C H A P T E R 1 6

Standardized Test Prep

Answer the following items on a separate piece of paper. 5.At a constant pressure, the following reaction is
exothermic: 2NO2(g) ⎯→ N2O4(g). Which of
MULTIPLE CHOICE the following statements is true about the reac-
tion (as written)?
1.Which of the following two conditions will favor
a spontaneous reaction? A. The reaction is always spontaneous.
A. an increase in entropy and a decrease in B. The reaction is spontaneous at low tempera-
enthalpy
B. an increase in entropy and an increase in tures, but not at high temperatures.
enthalpy
C. a decrease in entropy and a decrease in C. The reaction is spontaneous at high tempera-
enthalpy
D. a decrease in entropy and an increase in tures, but not at low temperatures.
enthalpy
D. The reaction is never spontaneous.

2.The gasification of coal is a method of produc- SHORT ANSWER
ing methane by the following reaction.
6.For a reaction in which a diatomic molecule dis-
C(s) + 2H2(g) ⎯→ CH4(g) ΔH = ? sociates into atoms, what are the signs of ΔS and
ΔH? Explain your answer. You may use the
Find ΔH by using the enthalpy changes in the reaction N2(g) ⎯→ 2N(g) as an example of
following combustion reactions such a reaction.

C(s) + O2(g) ⎯→ CO2(g) ΔH = −394 kJ EXTENDED RESPONSE

H2(g) + 1 O2(g) ⎯→ H2O(l ) ΔH = −286 kJ 7.For certain molecules, enthalpies of formation
2 can be determined from combustion data. Using
the diagram below, calculate the enthalpy of
CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O(l) formation of methane gas, CH4(g), and the
ΔH = −891 kJ enthalpies for two of the combustion reactions
listed below.

A. 75 kJ C. 1856 kJ Combustion reactions:
B. −75 kJ D. −1856 kJ
C(s) + O2(g) ⎯→ CO2(g) ΔH = −393.5 kJ

3.Two metals of equal mass but different H2(g) + 1 O2(g) ⎯→ H2O(l ) ΔH = ?
specific heats absorb the same amount of heat. 2
Which metal undergoes the smaller change in
temperature? CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O(l) ΔH = ?
A. The metal with the higher specific heat.
B. The metal with the lower specific heat. C(s) + 2H2(g) + 2O2(g)
C. Both undergo the same change in temperature.
D. Cannot determine from the information given. CH4(g) + 202(g)

4.Which of the following processes has a Enthalpy –965.1 kJ 890.2 kJ
negative ΔS? (kJ)
A. evaporating 1 mol of a liquid
B. raising the temperature of 1 L of water from CO2(g) + 2H2O(l)
295 K to 350 K
C. freezing of 1 mol of a liquid
D. None of the above

Always read the question before
looking at the answer choices.

R E A C T I O N E N E R G Y 557

CHAPTER LAB SEE PRE-LAB

CALORIMETRY

Calorimetry and Hess’s Law

OBJECTIVES BACKGROUND
• Demonstrate proficiency in the use of Hess’s law states that the overall enthalpy change in
a reaction is equal to the sum of the enthalpy changes
calorimeters and related equipment. in the individual steps in the process. In this experi-
• Relate temperature changes to enthalpy ment, you will use a calorimeter to measure the
energy released in three chemical reactions. From
changes. your experimental data, you will verify Hess’s law.
• Determine enthalpies of reaction for several
SAFETY
reactions.
• Demonstrate that enthalpies of reactions For review of safety, please see Safety in the
Chemistry Laboratory in the front of your book.
can be additive.
PREPARATION
MATERIALS 1. Prepare a data table in which to record the total
• 4 g NaOH pellets volumes of liquid, initial temperature, and final
• 50 mL 1.0 M HCl acid solution temperature of the three reactions you will carry
• 50 mL 1.0 M NaOH solution out, as well as the mass of the empty watch glass
• 100 mL 0.50 M HCl solution and the watch glass plus NaOH pellets.
• 100 mL graduated cylinder
• balance 2. Gently insert the thermometer into the plastic
• distilled water foam cup held upside down. Thermometers
• forceps break easily, so be careful with them, and do not
• glass stirring rod use them to stir a solution.
• gloves
• plastic-foam cups (or calorimeter) PROCEDURE
• spatula Reaction 1: Dissolving NaOH
• thermometer
• watch glass 1. Pour 100 mL of distilled water into your
calorimeter. Record the water temperature to
558 C H A P T E R 1 6 the nearest 0.1°C.

2. Weigh a clean and dry watch glass to the nearest
0.01 g. Wearing gloves and using forceps, place
about 2 g of NaOH pellets on the watch glass.
Measure and record the mass of the watch glass
and the pellets to the nearest 0.01 g. It is impor-
tant that this step be done quickly: NaOH
absorbs moisture from the air.

3. Immediately place the NaOH pellets in the CLEANUP AND DISPOSAL
calorimeter cup, and gently stir the solution 12. Check with your teacher for the proper
with a stirring rod. Do not stir with a ther-
mometer. Place the lid on the calorimeter. disposal procedures. Always wash your
Watch the thermometer, and record the highest hands thoroughly after cleaning up the
temperature in your data table. lab area and equipment.

4. Be sure to clean all equipment and rinse it with ANALYSIS AND INTERPRETATION
distilled water before continuing. 1. Organizing Ideas: Write a balanced chemical
equation for each of the three reactions that
Reaction 2: NaOH and HCI in Solution you performed. (Hint: Be sure to include the
5. Pour 50 mL of 1.0 M HCl into your calorimeter. physical states of matter for all substances.)
Record the temperature of the HCl solution to
the nearest 0.1°C. 2. Organizing Ideas: Write the equation for the
total reaction by adding two of the equations
6. Pour 50 mL of 1.0 M NaOH into a graduated from item 1 and then canceling out substances
cylinder. For this step only, rinse the thermome- that appear in the same form on both sides of
ter, and measure the temperature of the NaOH the new equation.
solution in the graduated cylinder to the nearest
0.1°C. Record the temperature, and then replace 3. Organizing Data: Calculate the change in
the thermometer in the calorimeter. temperature for each of the reactions.

7. Pour the NaOH solution into the calorimeter 4. Organizing Data: Assuming that the density of
cup, and stir gently. Place the lid on the calorim- the water and the solutions is 1.00 g/mL, calcu-
eter. Watch the thermometer and record the late the mass of liquid present for each of the
highest temperature. reactions.

8. Pour the solution in the container designated by 5. Analyzing Results: Using the calorimeter equa-
your teacher. Clean and rinse all equipment tion, calculate the energy as heat released by
before continuing with the procedure. each reaction. (Hint: Use the specific heat of
water in your calculations.)
Reaction 3: Solid NaOH and HCI in Solution
9. Pour 100 mL of 0.50 M HCl into your calorime- cp, H2O = 4.184 J/g • °C
ter. Record the temperature of the HCl solution heat = m × Δt × cp, H2O
to the nearest 0.1°C. 6. Organizing Data: Calculate the moles of NaOH
used in each of the reactions.
10. Measure the mass of a clean and dry watch
glass, and record the mass. Wear gloves, and 7. Analyzing Results: Calculate the ΔH value in
using forceps, obtain approximately 2 g of kJ/mol of NaOH for each of the three reactions.
NaOH pellets. Place them on the watch glass,
and record the total mass. As in step 2, it is 8. Organizing Ideas: Using your answer to
important that this step be done quickly. Analysis and Interpretation item 2 and your
knowledge of Hess’s law, explain how the
11. Immediately place the NaOH pellets in the enthalpies for the three reactions should be
calorimeter, and gently stir the solution. Place mathematically related.
the lid on the calorimeter. Watch the thermome-
ter, and record the highest temperature. When
finished with this reaction, pour the solution into
the container designated by your teacher for dis-
posal of basic solutions.

R E A C T I O N E N E R G Y 559

CHAPTER 17

Reaction Kinetics

Chemists can determine the rates
at which chemical reactions occur.

The Thermite Reaction

The Reaction Process SECTION 1

OBJECTIVES

B y studying many types of experiments, chemists have found that Explain the concept of
reaction mechanism.
chemical reactions occur at widely differing rates. For example, in the
presence of air, iron rusts very slowly, whereas the methane in natural Use the collision theory to
gas burns rapidly. The speed of a chemical reaction depends on the interpret chemical reactions.
energy pathway that a reaction follows and the changes that take place
on the molecular level when substances interact. In this chapter, you Define activated complex.
will study the factors that affect how fast chemical reactions take place.

Reaction Mechanisms Relate activation energy
to enthalpy of reaction.

If you mix aqueous solutions of HCl and NaOH, an extremely rapid
neutralization reaction occurs, as shown in Figure 1.

H3O+(aq) + Cl−(aq) + Na+(aq) + OH−(aq) ⎯→ 2H2O(l) + Na+(aq) + Cl−(aq)

The reaction is practically instantaneous; the rate is limited only by the
speed with which the H3O+ and OH− ions can diffuse through the water
to meet each other. On the other hand, reactions between ions of the
same charge and between molecular substances are not instantaneous.
Negative ions repel each other, as do positive ions. The electron clouds
of molecules also repel each other strongly at very short distances.
Therefore, only ions or molecules with very high kinetic energy can
overcome repulsive forces and get close enough to react. In this section,
we will limit our discussion to reactions between molecules.

Colorless hydrogen gas consists of pairs of hydrogen atoms bonded
together as diatomic molecules, H2. Violet-colored iodine vapor is also
diatomic, consisting of pairs of iodine atoms bonded together as I2 mol-
ecules. A chemical reaction between these two gases at elevated tem-
peratures produces hydrogen iodide, HI, a colorless gas. Hydrogen
iodide molecules, in turn, tend to decompose and re-form hydrogen
and iodine molecules, producing the violet gas shown in Figure 2. The
following chemical equations describe these two reactions.

H2(g) + I2(g) ⎯→ 2HI(g)

2HI(g) ⎯→ H2(g) + I2(g) FIGURE 1 As NaOH solution is
poured into HCl solution, a very
Such equations indicate only which molecular species disappear as a rapid neutralization reaction occurs.
result of the reactions and which species are produced. They do not Excess NaOH turns the phenolph-
show the reaction mechanism, the step-by-step sequence of reactions by thalein indicator pink.
which the overall chemical change occurs.

R E A C T I O N K I N E T I C S 561

FIGURE 2 Colorless hydrogen Although only the net chemical change is directly observable for
iodide gas, HI, decomposes into col- most chemical reactions, experiments can often be designed that sug-
orless hydrogen gas and violet gest the probable sequence of steps in a reaction mechanism. Each reac-
iodine gas. Some of the hydrogen gas tion step is usually a simple process. The equation for each step
and iodine gas will reform HI. represents the actual atoms, ions, or molecules that participate in that
step. Even a reaction that appears from its balanced equation to be a
simple process may actually be the result of several simple steps.

For many years, the formation of hydrogen iodide, as shown in Figure
2, was considered a simple one-step process. It was thought to involve the
interaction of two molecules, H2 and I2, in the forward reaction and two
HI molecules in the reverse reaction. Experiments eventually showed,
however, that a direct reaction between H2 and I2 does not take place.

Alternative mechanisms for the reaction were proposed based on
the experimental results. The steps in each reaction mechanism had to
add together to give the overall equation. Note that two of the species
in the mechanism steps—I and H2I—do not appear in the net equation.
Species that appear in some steps but not in the net equation are known
as intermediates. (Notice that they cancel each other out in the follow-
ing mechanisms.) The first possible mechanism has the following two-
step pathway.

Step 1: I2 ←⎯⎯→ 2I
Step 2: ←⎯⎯→
2I + H2 ←⎯⎯→ 2HI
I2 + H2 2HI

The second possible mechanism has a three-step pathway.

Step 1: I+ I2 ←⎯⎯→ 2I
Step 2: H2 ←⎯⎯→
Step 3: H2I +I ←⎯⎯→ H2I
I2 + ←⎯⎯→ 2HI
H2
2HI

The reaction between hydrogen gas and iodine vapor to produce hydro-
gen iodide gas is an example of a homogeneous reaction, a reaction
whose reactants and products exist in a single phase—in this case, the gas
phase.This reaction system is also an example of a homogeneous chemi-
cal system because all reactants and products in all intermediate steps
are in the same phase.

Collision Theory

In order for reactions to occur between substances, their particles (mol-
ecules, atoms, or ions) must collide. Furthermore, these collisions must
result in interactions. The set of assumptions regarding collisions and
reactions is known as collision theory. Chemists use this theory to inter-
pret many of their observations about chemical reactions.

562 C H A P T E R 1 7

AB AB AB AB AB AB

BB AB AB AB AB
(c) Collision that has
A2 (b) Collision poor orientation
(a) Effective collision, too gentle
FIGURE 3 Three possible colli-
favorable orientation and energy sion patterns for AB molecules are
shown. Not every collision produces
Consider what might happen on a molecular scale in one step of a a chemical reaction.
homogeneous reaction system. We will analyze a proposed first step in
a hypothetical decomposition reaction. R E A C T I O N K I N E T I C S 563

AB + AB ←⎯⎯→ A2 + 2B

According to the collision theory, the two AB molecules must collide in
order to react. Furthermore, they must collide with a favorable orienta-
tion and with enough energy to merge the valence electrons and disrupt
the bonds of the AB molecules. If they do so, a reshuffling of bonds
leads to the formation of the products, one A2 molecule and two B
atoms. An effective collision is modeled in Figure 3a.

If a collision is too gentle, the two molecules simply rebound from each
other unchanged. This effect is illustrated in Figure 3b. Similarly, a col-
lision in which the reactant molecules have an unfavorable orientation
has little effect. The colliding molecules rebound without reacting. A
collision that has poor orientation is shown in Figure 3c.

A chemical reaction produces new bonds which are formed between
specific atoms in the colliding molecules. Unless the collision brings the
correct atoms close together and in the proper orientation, the molecules
will not react. For example, if a chlorine molecule collides with the oxygen
end of a nitrogen monoxide molecule, the following reaction may occur.

NO(g) + Cl2(g) ⎯→ NOCl(g) + Cl(g)

This reaction will not occur if the chlorine molecule strikes the nitro-
gen end of the molecule.

Thus, collision theory provides two reasons why a collision between
reactant molecules may fail to produce a new chemical species: the col-
lision is not energetic enough to supply the required energy, or the col-
liding molecules are not oriented in a way that enables them to react
with each other.

Activation Energy

Consider the reaction for the formation of water from the diatomic
gases oxygen and hydrogen according to the following equation.

2H2(g) + O2(g) ⎯→ 2H2O(l)

The enthalpy of formation is quite high: ΔH 0 = −285.8 kJ/mol at
f
298.15 K. The free-energy change is also large: ΔG0 = −237.1 kJ/mol.

Why, then, don’t oxygen and hydrogen combine spontaneously and

immediately to form water when they are mixed at room temperature?

Hydrogen and oxygen gases exist as diatomic molecules. When the

molecules approach each other, the electron clouds repel each other, so

the molecules might not meet. For a reaction to occur, the colliding mol-

ecules must have enough kinetic energy to intermingle the valence elec-

trons. In other words, the bonds of these molecular species must be

broken in order for new bonds to be formed between oxygen and hydro-

gen atoms. Bond breaking is an endothermic process, and bond forming

is exothermic. Even though the net process for forming water is exother-

mic, an initial input of energy is needed to overcome the repulsion forces

that occur between reactant molecules when they are brought very close

together. This initial energy input activates the reaction.

Once an exothermic reaction is started, the energy released is

enough to sustain the reaction by activating other molecules. Thus, the

reaction rate keeps increasing. It is limited only by the time required for

reactant particles to acquire the energy and make

contact. Energy from an outside source may start

Reaction Pathways for Forward exothermic reactants along the pathway of reac-
and Reverse Reactions tion. A generalized reaction pathway for an
exothermic reaction is shown as the forward reac-

Activated complex tion in Figure 4. The minimum amount of energy
needed to activate this reaction is the activation

Ea energy represented by Ea. Activation energy is the
minimum energy required to transform the reactants

Energy Reactants Ea′ into an activated complex.
ΔE The reverse reaction, decomposition of water
Products
molecules, is endothermic because the water mole-
cules lie at an energy level lower than that of the

Forward Reverse hydrogen and oxygen molecules. The water mole-
reaction reaction cules require a larger activation energy before they

(exothermic) (endothermic) can decompose to re-form oxygen and hydrogen.

The energy needed to activate an endothermic

Course of reaction reaction is greater than that required for the origi-

FIGURE 4 The difference between the activation energies nal exothermic change and is represented by Ea′ in
for the reverse and forward reactions of a reversible reaction Figure 4. The difference between Ea′ and Ea is
equals the energy change in the reaction, ΔE. The quantity for equal to the energy change in the reaction, ΔE. This
ΔE is the same for both directions, but is negative for the energy change has the same numerical value for the
exothermic direction and positive for the endothermic
direction. forward reaction as it has for the reverse reaction

but with the opposite sign.

564 C H A P T E R 1 7

The Activated Complex

When molecules collide, some of their high kinetic energy is converted

into internal potential energy within the colliding molecules. If enough

energy is converted, molecules with suitable orientation become acti-

vated. New bonds can then form. In this brief interval of bond breakage

and bond formation, the collision complex is in a transition state. Some

partial bonding exists in this transitional structure. A transitional struc-

ture that results from an effective collision and that persists while old

bonds are breaking and new bonds are forming is called an activated

complex.

Figure 5 graphically breaks down the reaction pathway of the for-

mation of hydrogen iodide gas into three steps. Beginning with the reac-

tants, H2 and I2, a certain amount of activation energy, Ea1, is needed to
form the activated complex that leads to the formation of the interme-

diates H2 and 2I. Then more activation energy, Ea2, is needed to form
the activated complex leading to the intermediates H2I and I. In order
to arrive at the final product, 2HI, another increase in activation energy

is necessary, as seen by the highest peak labeled Ea3.
An activated complex is formed when an effective collision raises the

internal energies of the reactants to their minimum level for reaction, as

in Figure 4. Both forward and reverse reactions go through the same acti-

vated complex. A bond that is broken in the activated complex for the

forward reaction must be re-formed in the activated complex for the

reverse reaction. Observe that an activated complex occurs at a high-

energy position along the reaction pathway.

The kinetic-molecular theory states that the speeds and therefore the

kinetic energies of the molecules increase as the

temperature increases. An increase in speed causes Activation Energy Peaks
more collisions, which can cause an increase in the in the Formation of Activated Complexes
number of reactions. However, an increase in the

reaction rate depends on more than simply the num- *
ber of collisions, as Figure 3 illustrates.The collisions

between molecules must possess sufficient energy to

form an activated complex or a reaction will not Ea3
take place. Raising the temperature of a reaction

provides more molecules that have this activation * E*a2
energy and causes an increase in the reaction rate.
Energy
In its brief existence, the activated complex H2 H2I
has partial bonding that is characteristic of both Ea1+ 2I +I
reactant and product. It may then re-form the

original bonds and separate back into the reac- H2 + I2 ΔE
tant particles, or it may form new bonds and sep- 2HI
arate into product particles. The activated

complex, unlike the relatively stable intermediate Course of reaction

products, is a very short-lived molecular complex FIGURE 5 This energy profile graphically shows the forma-
in which bonds are in the process of being broken tion of three activated complexes (*) during the gas-phase
and formed. reaction H2 + I2 ⎯→ 2HI.

R E A C T I O N K I N E T I C S 565

SAMPLE PROBLEM A For more help, go to the Math Tutor at the end of this chapter.

Copy the energy diagram below, and label the reactants, products, ⌬E, Ea, and Ea′. Determine the value
of ⌬Eforward , ⌬Ereverse , Ea , and Ea′.

80

Energy (kJ/mol) 50

0

Forward Reverse

SOLUTION The energy level of reactants is always at the left-hand end of such a curve, and the energy
level of products is always at the right-hand end. The energy change in the reaction, ΔE, is the
difference between these two energy levels. The activation energy differs in the forward and

reverse directions. It is the minimum energy needed to achieve effective reaction in either

direction. As Ea, it is the difference between the reactant energy level and the peak in the
curve. As Ea′, it is the difference between the product energy level and the peak in the curve.

Energy (kJ/mol) 80 Ea′
Products
50
Ea ΔE

0
Reactants

Forward Reverse

ΔEforward = energy of products − energy of reactants
ΔEforward = 50 kJ/mol − 0 kJ/mol = +50 kJ/mol

ΔEreverse = energy of reactants − energy of products
ΔEreverse = 0 kJ/mol − 50 kJ/mol = − 50 kJ/mol

Ea = energy of activated complex − energy of reactants
Ea = 80 kJ/mol − 0 kJ/mol = 80 kJ/mol

Ea′ = energy of activated complex − energy of products
Ea′ = 80 kJ/mol − 50 kJ/mol = 30 kJ/mol

566 C H A P T E R 1 7

PRACTICE Answers in Appendix E

1. a. Use the method shown in the sample problem to redraw and
label the following energy diagram. Determine the value of
ΔEforward, ΔEreverse, Ea, and Ea′.

b. Is the forward reaction shown in the diagram exothermic or
endothermic? Explain your answer.

Energy (kJ/mol) 100
50
0

−50
−100
−150

Forward Reverse

2. a. Draw and label an energy diagram similar to the one shown in Go to go.hrw.com for
the sample problem for a reaction in which Ea = 125 kJ/mol and more practice problems
Ea′ = 86 kJ/mol. Place the reactants at energy level zero. that ask you to
determine Ea and ΔE.
b. Calculate the values of ΔEforward and ΔEreverse.
Keyword: HC6RXKX
c. Is this reaction endothermic or exothermic? Explain your
answer.

3. a. Draw and label an energy diagram for a reaction in which
Ea = 154 kJ/mol and ΔE = 136 kJ/mol.

b. Calculate the activation energy, Ea′, for the reverse reaction.

SECTION REVIEW 6. What is the difference between an activated com-
plex and an intermediate?
1. What is meant by reaction mechanism?
2. What factors determine whether a molecular colli- 7. Explain why, even though a collision may have
energy in excess of the activation energy, reaction
sion produces a reaction? may not occur.
3. What is activation energy?
4. What is an activated complex? Critical Thinking
5. How is activation energy related to the energy
8. ANALYZING INFORMATION Which corresponds to
of reaction? the faster rate: a mechanism with a small activa-
tion energy or one with a large activation energy?
Explain your answer.

R E A C T I O N K I N E T I C S 567

SECTION 2 Reaction Rate

OBJECTIVES T he change in concentration of reactants per unit time as a reaction

Define chemical kinetics, and proceeds is called the reaction rate. The study of reaction rates is con-
explain the two conditions cerned with the factors that affect the rate and with the mathematical
necessary for chemical expressions that reveal the specific dependencies of the rate on concen-
reactions to occur. tration. The area of chemistry that is concerned with reaction rates and
Discuss the factors that influ- reaction mechanisms is called chemical kinetics.
ence reaction rate.
Define catalyst, and discuss Rate-Influencing Factors
two different types.
Relate the order of a reaction For reactions other than simple decompositions to occur, particles must
to the rate law for the come into contact in a favorable orientation and with enough energy for
reaction. activation. Thus, the rate of a reaction depends on the collision fre-
Explain and write rate laws quency of the reactants and on the collision efficiency. Any change in
for chemical reactions. reaction conditions that affects the collision frequency, the collision effi-
ciency, or the collision energy affects the reaction rate. At least five
www.scilinks.org important factors influence the rate of a chemical reaction.
Topic: Factors Affecting
Nature of Reactants
Reaction Rates
Code: HC60564 Substances vary greatly in their tendencies to react. For example, hydro-
gen combines vigorously with chlorine under certain conditions. Under
568 C H A P T E R 1 7 the same conditions, it may react only weakly with nitrogen. Sodium
and oxygen combine much more rapidly than iron and oxygen under
similar conditions. Bonds are broken and other bonds are formed in
reactions. The rate of reaction depends on the particular reactants and
bonds involved.

Surface Area

Gaseous mixtures and dissolved particles can mix and collide freely;
therefore, reactions involving them can occur rapidly. In heterogeneous
reactions, the reaction rate depends on the area of contact of the reac-
tion substances. Heterogeneous reactions involve reactants in two dif-
ferent phases. These reactions can occur only when the two phases are
in contact. Thus, the surface area of a solid reactant is an important fac-
tor in determining rate. An increase in surface area increases the rate of
heterogeneous reactions.

Solid zinc reacts with aqueous hydrochloric acid to produce zinc
chloride and hydrogen gas according to the following equation.

Zn(s) + 2HCl(aq) ⎯→ ZnCl2(aq) + H2(g)

This reaction occurs at the surface of the zinc solid. A cube of zinc mea-
suring 1 cm on each edge presents only 6 cm2 of contact area. The same
amount of zinc in the form of a fine powder might provide a contact
area thousands of times greater than the original area. Consequently,
the reaction rate of the powdered solid is much faster.

A lump of coal burns slowly when kindled in air. The rate of burning
can be increased by breaking the lump into smaller pieces, exposing
more surface area. If the piece of coal is powdered and then ignited
while suspended in air, it burns explosively. This is the cause of some
explosions in coal mines.

Temperature (a)

An increase in temperature increases the average kinetic energy of the
particles in a substance; this can result in a greater number of effective
collisions when the substance is allowed to react with another sub-
stance. If the number of effective collisions increases, the reaction rate
will increase.

To be effective, the energy of the collisions must be equal to or
greater than the activation energy. At higher temperatures, more parti-
cles possess enough energy to form the activated complex when colli-
sions occur. Thus, a rise in temperature produces an increase in collision
energy as well as in collision frequency.

Decreasing the temperature of a reaction system has the opposite
effect. The average kinetic energy of the particles decreases, so they col-
lide less frequently and with less energy, producing fewer effective col-
lisions. Beginning near room temperature, the reaction rates of many
common reactions roughly double with each 10 K (10°C) rise in tem-
perature. This rule of thumb should be used with caution, however. The
actual rate increase with a given rise in temperature must be deter-
mined experimentally.

Concentration (b)
FIGURE 6 Carbon burns faster in
Pure oxygen has five times the concentration of oxygen molecules that pure oxygen (a) than in air (b)
air has at the same pressure; consequently, a substance that oxidizes in because the concentration of the
air oxidizes more vigorously in pure oxygen. For example, in Figure 6, reacting species, O2, is greater.
the light produced when the lump of charcoal is burned in pure oxygen
is much more intense than the light produced when the charcoal lump R E A C T I O N K I N E T I C S 569
is heated in air until combustion begins. The oxidation of charcoal is a
heterogeneous reaction system in which one reactant is a gas. The reac-
tion rate depends not only on the amount of exposed charcoal surface
but also on the concentration of the reacting species, O2.

In homogeneous reaction systems, reaction rates depend on the con-
centration of the reactants. Predicting the mathematical relationship
between rate and concentration is difficult because most chemical reac-
tions occur in a series of steps, and only one of these steps determines
the reaction rate. If the number of effective collisions increases, the rate

(a) (b) (c) (d)

FIGURE 7 The number of mole- increases as well. In general, an increase in rate is expected if the con-
cules of reacting species affects the centration of one or more of the reactants is increased, as depicted by
number of possible collisions and the model in Figure 7. In the system with only two molecules, shown in
therefore the reaction rate. Figure 7a, only one collision can possibly occur. When there are four
molecules in the system, as in Figure 7b, there can be four possible col-
lisions. Under constant conditions, as the number of molecules in the
system increases, so does the total number of possible collisions
between them. Figure 7c and d show a five- and eight-molecule system,
allowing six and sixteen possible collisions, respectively. Lowering the
concentration should have the opposite effect. The actual effect of con-
centration changes on reaction rate, however, must be determined
experimentally.

FIGURE 8 The reaction rate of Presence of Catalysts
the decomposition of hydrogen per-
oxide, H2O2, can be increased by Some chemical reactions proceed quite slowly. Sometimes their reac-
using a catalyst. The catalyst used tion rates can be increased dramatically by the presence of a catalyst. A
here is manganese dioxide, MnO2, a catalyst is a substance that changes the rate of a chemical reaction with-
black solid. A 30% H2O2 solution is out itself being permanently consumed. The action of a catalyst is called
added dropwise onto the MnO2 in catalysis. The catalysis of the decomposition reaction of hydrogen per-
the beaker and rapidly decomposes oxide by manganese dioxide is shown in Figure 8. A catalyst provides an
to O2 and H2O. Both the oxygen and alternative energy pathway or reaction mechanism in which the
water appear as gases because the potential-energy barrier between reactants and products is lowered.
energy released by the reaction caus- The catalyst may be effective in forming an alternative activated com-
es much of the water to vaporize. plex that requires a lower activation energy—as suggested in the ener-
gy profiles of the decomposition of hydrogen peroxide, H2O2, shown in
570 C H A P T E R 1 7 Figure 9—according to the following equation.

2H2O2(l) ⎯→ O2(g) + 2H2O(l)

Catalysts do not appear among the final products of reactions they
accelerate. They may participate in one step along a reaction pathway
and be regenerated in a later step. In large-scale and cost-sensitive reac-
tion systems, catalysts are recovered and reused. A catalyst that is in the
same phase as all the reactants and products in a reaction system is called
a homogeneous catalyst. When its phase is different from that of the reac-
tants, it is called a heterogeneous catalyst. Metals are often used as
heterogeneous catalysts. The catalysis of many reactions is promoted by
adsorption of reactants on the metal surfaces, which has the effect of
increasing the concentration of the reactants.

Comparison of Pathways for the Decomposition FIGURE 9 The activation energy
of H2O2 by Various Catalysts for a chemical reaction can be
reduced by adding an appropriate
catalyst.

Energy (kJ/mol) Ea = 75 Uncatalyzed
Ea = 58
Ea = 4 Chemically Homogeneous
catalyzed (I-) processes
Enzyme
catalyzed

Heterogeneous
process (MnO2)

Course of reaction

Rate Laws for Reactions Chemistry in Action

The relationship between the rate of a reaction and the concentration Go to go.hrw.com for a full-length
of one reactant is determined experimentally by first keeping the con- article on using spectroscopy to
centrations of other reactants and the temperature of the system con- measure reaction kinetics.
stant. Then the reaction rate is measured for various concentrations of
the reactant in question. A series of such experiments reveals how the Keyword: HC6RXKX
concentration of each reactant affects the reaction rate.
R E A C T I O N K I N E T I C S 571
Hydrogen gas reacts with nitrogen monoxide gas at constant volume
and at an elevated constant temperature, according to the following
equation.

2H2(g) + 2NO(g) ⎯→ N2(g) + 2H2O(g)

Four moles of reactant gases produce three moles of product gases;
thus, the pressure of the system diminishes as the reaction proceeds. The
rate of the reaction can, therefore, be determined by measuring the
change of pressure in the vessel with time.

Suppose a series of experiments is conducted using the same initial
concentration of nitrogen monoxide but different initial concentrations
of hydrogen. The initial reaction rate is found to vary directly with the
hydrogen concentration: doubling the concentration of H2 doubles the
rate, and tripling the concentration of H2 triples the rate. If R represents
the reaction rate and [H2] is the concentration of hydrogen in moles per

Explosives liter, the mathematical relationship between rate and concentration can
be expressed as follows.
In a tiny fraction of a second, the reac-
tions of explosives such as nitroglyc- R ∝ [H2]
erin, trinitrotoluene (TNT), and
dynamite are over. These materials are The ∝ is a symbol that is read “is proportional to.”
primarily organic substances containing Now suppose the same initial concentration of hydrogen is used but
mostly carbon, hydrogen, oxygen, and
nitrogen atoms held together by rela- the initial concentration of nitrogen monoxide is varied.The initial reac-
tively weak bonds. When “set off,” tion rate is found to increase fourfold when the NO concentration is
explosive materials experience rapid doubled and ninefold when the concentration of NO is tripled. Thus, the
decomposition. The released elements reaction rate varies directly with the square of the nitrogen monoxide
immediately react to form gaseous N2, concentration, as described by the following proportion.
CO, CO2, and NO2. The bonds in these
small molecules are much stronger R ∝ [NO]2
than those in the original explosive
material, and so an enormous amount Because R is proportional to [H2] and to [NO]2, it is proportional to
of energy is released. In addition, the their product.
sudden formation of gaseous material
causes a tremendous increase in pres- R ∝ [H2][NO]2
sure that provides the force to demol-
ish an unwanted building or break By introduction of an appropriate proportionality constant, k, the
rocks for building roads. expression becomes an equality.

572 C H A P T E R 1 7 R = k[H2][NO]2

An equation that relates reaction rate and concentrations of reactants is
called the rate law for the reaction. It is applicable for a specific reaction
at a given temperature. A rise in temperature increases the reaction
rates of most reactions. The value of k usually increases as the tempera-
ture increases, but the relationship between reaction rate and concen-
tration almost always remains unchanged.

Using the Rate Law

The general form for the rate law is given by the following equation:

R = k[A]n[B]m

The reaction rate is represented by R, k is the specific rate constant,
and [A] and [B] represent the molar concentrations of reactants. The
respective powers to which the concentrations are raised are represented
by n and m. The rate law is applicable for a specific reaction at a given set
of conditions and must be determined from experimental data.

The power to which a reactant concentration is raised is called the
order in that reactant. The value of n is said to be the order of the reac-
tion with respect to [A], so the reaction is said to be “nth order in A.”
Similarly, for the value of m, the reaction is said to be “mth order in B.”
The orders, or powers, n and m, are usually small integers or zero. An
order of one for a reactant means that the reaction rate is directly pro-
portional to the concentration of that reactant. An order of two means

that the reaction rate is directly proportional to the square of the reac-
tant. An order of zero means that the rate does not depend on the con-
centration of the reactant, as long as some of the reactant is present. The
sum of all of the reactant orders is called the order of the reaction, or
overall order. The overall order of the reaction is equal to the sum of the
reactant orders, or n + m. Some examples of observed rate laws that
have been derived experimentally are shown below. Some of these reac-
tions involve nitrogen oxides, which are highly reactive gases that con-
tribute to the formation of smog that can blanket an entire city, as
shown in Figure 10.

3NO(g) ⎯→ N2O(g) + NO2(g) R = k[NO]2
second order in NO,
second order overall

NO2(g) + CO(g) ⎯→ NO(g) + CO2(g) R = k[NO2]2
second order in NO2,
zero order in CO,

second order overall

2NO2(g) ⎯→ 2NO(g) + O2(g) R = k[NO2]2
second order in NO2,
second order overall

2H2O2(aq) ⎯→ 2H2O(l) + O2(g) R = k[H2O2]
first order in H2O2,
first order overall

It is important to understand that the orders in the rate law may or may
not match the coefficients in the balanced equation. These orders must
be determined from experimental data.

Specific Rate Constant FIGURE 10 A cloud of polluted
air, commonly known as smog, set-
The specific rate constant (k) is the proportionality constant relating the tles over a city. Smog is common in
rate of the reaction to the concentrations of reactants. It is important to industrialized areas, where highly
remember the following about the value of k: reactive gases and particulate matter
are released into the air.
1. Once the reaction orders (powers) are known, the value of k must be
determined from experimental data. R E A C T I O N K I N E T I C S 573

2. The value of k is for a specific reaction; k has a different value for
other reactions, even at the same conditions.

3. The units of k depend on the overall order of the reaction.
4. The value of k does not change for different concentrations of reac-

tants or products. So, the value of k for a reaction remains the same
throughout the reaction and does not change with time.
5. The value of k is for the reaction at a specific temperature; if we
increase the temperature of the reaction, the value of k increases.
6. The value of k changes (becomes larger) if a catalyst is present.

SAMPLE PROBLEM B For more help, go to the Math Tutor at the end of this chapter.

Three experiments that have identical conditions were performed to measure the initial rate of
the reaction

2HI(g) ⎯→ H2(g) + I2(g)
The results for the three experiments in which only the HI concentration was varied are as follows:

Experiment [HI] (M) Rate (M/s)
1 0.015 1.1 × 10−3
2 0.030 4.4 × 10−3
3 0.045 9.9 × 10−3

Write the rate law for the reaction. Find the value and units of the specific rate constant.

SOLUTION

1 ANALYZE The general rate law for this reaction has the form R = k[HI]n. We need to deduce the value
of the power n.

2 PLAN Find the ratio of the reactant concentrations between two experiments, such as 1 and 2, [[⎯HH⎯II]]21.
3 COMPUTE Then, see how the ratio of concentration affects the ratio of rates, ⎯RR⎯21.

Concentration ratio: [⎯HI]2 = 0⎯.030 M = 2.0; rate ratio: R⎯2 = ⎯41..14 ×× 1100⎯−−33 MM//ss = 4.0
[HI]1 0.015 M R1

Thus, when the concentration changes by a factor of 2, the rate changes by a factor of 4, or
22, so the rate law is R = k[HI]2.

To find the value of k, we can rearrange the rate law and substitute known values for any
one experiment. Do the following for Experiment 1:

k = [⎯HRI]2 = 1⎯.(10×.01150⎯−M3 M)2/s = 4.9 M−1s−1

4 EVALUATE By comparing items 1 and 3 in the table, we see that when [HI] is tripled, the rate changes
by a factor of 9, or 32. This rate change confirms that the order is 2. The same value of k can

be calculated from any other experiment. Thus, the rate law and k are correct.

PRACTICE Answers in Appendix E

1. For the reaction 3A ⎯→ C, the initial concentration of A was 0.2 Go to go.hrw.com for
M and the reaction rate was 1.0 M/s. When [A] was doubled, the more practice problems
reaction rate increased to 4.0 M/s. Determine the rate law for the that ask you to
reaction. determine rate law and
rate constant.
2. The rate law for a reaction is found to be rate = k[X]3. By what
factor does the rate increase if [X] is tripled? Keyword: HC6RXKX

574 C H A P T E R 1 7

SAMPLE PROBLEM C For more help, go to the Math Tutor at the end of this chapter.

Three experiments were performed to measure the initial rate of the reaction

A + B ⎯→ C

Conditions were identical in the three experiments, except that the concentrations of reactants varied. The
results are as follows:

Experiment [A] (M) [B] (M) Rate (M/s)
1 1.2 2.4 8.0 × 10−8
2 1.2 1.2 4.0 × 10−8
3 3.6 2.4 7.2 × 10−7

Write the rate law for the reaction. Find the value and units of the specific rate constant.

SOLUTION

1 ANALYZE The general rate law for this reaction has the form R = k[A]n[B]m. We need to calculate the
values of the powers n and m.

2 PLAN Find the ratio of the reactant concentrations between two experiments that have the same
reaxtpees,riR⎯Rm⎯21; ethnitss
[A] but different [B]. Then, see how this ratio affects the ratio of ratio of rates
lets us find the value of m. A similar approach of comparing two that have the

same [B] but a different [A] lets us find the value of n.

3 COMPUTE First compare Experiments 1 and 2, which have the same [A], to find m:

Concentration ratio: [⎯B]1 = 2⎯.4 M = 2.0; rate ratio: R⎯1 = 4⎯8..00 ×× 1100⎯−−88 MM//ss = 2.0
[B]2 1.2 M R2

Thus, when the concentration of B changes by a factor of 2, the rate changes by a factor of
2, or 21. So, m is 1, and the reaction is first order in B.

Then, compare Experiments 1 and 3, which have the same [B], to find n:

Concentration ratio: [⎯A]3 = 3⎯.6 M = 3.0; rate ratio: R⎯3 = 7⎯8..20 ×× 1100⎯−−78 MM//ss = 9.0
[A]1 1.2 M R1

Thus, when the concentration of A changes by a factor of 3, the rate changes by a factor of
9, or 32. So, n is 2, and the reaction is second order in A.

The rate law is R ϭ k[A]2[B].

To find the value of k, we can rearrange the rate law and substitute known values for any
one experiment. Do the following for Experiment 1:

k = [⎯A]R2[B] = (⎯18..20 M× )120⎯(−28.M4 M/s ) = 2.3 × 10−8 M−2s−1

4 EVALUATE The same value of k can be calculated from the data for any other experiment. So, the rate
law and the calculation of k are correct.

R E A C T I O N K I N E T I C S 575

Rate Laws and Reaction Pathway

The form of the rate law depends on the reaction mechanism. For a
reaction that occurs in a single step, the reaction rate of that step is
proportional to the product of the reactant concentrations, each of
which is raised to its stoichiometric coefficient. For example, suppose
one molecule of gas A collides with one molecule of gas B to form two
molecules of substance C, according to the following equation.

A + B ⎯→ 2C

One particle of each reactant is involved in each collision. Thus, dou-
bling the concentration of either reactant will double the collision fre-
quency. It will also double the reaction rate for this step. Therefore, the
rate for this step is directly proportional to the concentration of A and
B. The rate law for this one-step forward reaction follows.

Rforward = kforward[A][B]

Now suppose the reaction is reversible. In the reverse step, two mol-
ecules of C must decompose to form one molecule of A and one of B or
2C ⎯→ A + B.

Thus, the reaction rate for this reverse step is directly proportional to
[C] × [C]. The rate law for the reverse step is Rreverse = kreverse[C]2.

The power to which the molar concentration of each reactant is raised
in the rate laws above corresponds to the coefficient for the reactant in
the balanced chemical equation. Such a relationship holds only if the
reaction follows a simple one-step path, that is, if the reaction occurs at
the molecular level exactly as written in the chemical equation.

If a chemical reaction proceeds in a sequence of steps, the rate law is
determined from the slowest step because it has the lowest rate. This
slowest-rate step is called the rate-determining step for the chemical
reaction.

Consider the reaction of nitrogen dioxide and carbon monoxide.

FIGURE 11 This diagram is a NO2(g) + CO(g) ⎯→ NO(g) + CO2(g)
representation of Step 1 in the reac-
tion of nitrogen dioxide and carbon The reaction is believed to be a two-step process represented by the fol-
monoxide. Notice the formation of lowing mechanism.
the intermediate species NO3 after
two molecules of NO2 collide. Step 1: NO2 + NO2 ⎯→ NO3 + NO slow
Step 2: NO3 + CO ⎯→ NO2 + CO2 fast
2NO2 NO3 + NO
In the first step, shown in Figure 11, two molecules of NO2
collide, forming the intermediate species NO3. This molecule
then collides with one molecule of CO and reacts quickly to
produce one molecule each of NO2 and CO2. The first step is
the slower of the two and is therefore the rate-determining
step. We can write the rate law from this rate-determining
step, which has two molecules of NO2 as the reactants.

R = k [NO2]2

The rate law does not include [CO] because CO reacts after
the rate-determining step and does not affect the rate.

576 C H A P T E R 1 7

SAMPLE PROBLEM D For more help, go to the Math Tutor at the end of this chapter.

Nitrogen dioxide and fluorine react in the gas phase according to the following equation.

2NO2(g) + F2(g) ⎯→ 2NO2F(g)
A proposed mechanism for this reaction follows.

Step 1: NO2 + F2 ⎯→ NO2F + F (slow)
Step 2: F + NO2 ⎯→ NO2F (fast)

Identify the rate-determining step, and write an acceptable rate law.

SOLUTION If we combine these two steps, the intermediate, F, cancels out and we are left with
the original equation. The first step is the slower step, and is considered the rate-
determining step. We can write the rate law from this rate-determining step.

R = k [NO2][F2]

SAMPLE PROBLEM E

A reaction involving reactants X and Y was found to occur by a one-step mechanism:
X + 2Y ⎯→ XY2. Write the rate law for this reaction, and then determine the effect
of each of the following on the reaction rate:
a. doubling the concentration of X
b. doubling the concentration of Y
c. using one-third the concentration of Y

SOLUTION Because the equation represents a single-step mechanism, the rate law can be written

from the equation (otherwise, it could not be). The rate will vary directly with the concentra-

tion of X, which has an implied coefficient of 1 in the equation. And the rate will vary direct-
ly with the square of the concentration of Y, which has the coefficient of 2: R = k[X][Y]2.

a. Doubling the concentration of X will double the rate (R = k[2X][Y]2).

b. Doubling the concentration of Y will increase the rate fourfold (R = k[X][2Y]2).

c. Using one-third the concentration of Y will reduce the rate to one-ninth of its original

value (R = k[X][ 1 Y]2).
3

PRACTICE Answers in Appendix E

1. The rate of a hypothetical reaction involving L and M is found to Go to go.hrw.com for
double when the concentration of L is doubled and to increase more practice problems
fourfold when the concentration of M is doubled. Write the rate that ask you to
law for this reaction. determine rate laws and
effects on reaction rates.
2. At temperatures below 498 K, the following reaction takes place.
Keyword: HC6RXKX
NO2(g) + CO(g) ⎯→ CO2(g) + NO(g)

Doubling the concentration of NO2 quadruples the rate of CO2
formed if the CO concentration is held constant. However, dou-
bling the concentration of CO has no effect on the rate of CO2
formation. Write a rate-law expression for this reaction.

R E A C T I O N K I N E T I C S 577

Wear safety goggles and an apron.

Factors Influencing Reaction Rate Materials

Question 3. To one test tube, add 10 mL of • Bunsen burner
• paper ash
How do the type of reactants, vinegar; to a second, add 5 mL • copper foil strip
surface area of reactants, of vinegar plus 5 mL of water; • graduated cylinder, 10 mL
concentration of reactants, and to a third, add 2.5 mL of • magnesium ribbon
and catalysts affect the rates vinegar plus 7.5 mL of water. • matches
of chemical reactions? To each of the three test tubes, • paper clip
add a 3 cm piece of magne- • sandpaper
Procedure sium ribbon. • steel wool
• 2 sugar cubes
Remove all combustible material 4. Using tongs, hold a sugar cube • white vinegar
from the work area. Wear safety • zinc strip
goggles and an apron. Record all and try to ignite it with a • 6 test tubes, 16 × 150 mm
your results in a data table. match. Then try to ignite it in a • tongs
burner flame. Rub paper ash
1. Add 10 mL of vinegar to each on a second cube, and try to
ignite it with a match.
of three test tubes. To one test
tube, add a 3 cm piece of Discussion
magnesium ribbon; to a sec-
ond, add a 3 cm zinc strip; and 1. What are the rate-influencing
to a third, add a 3 cm copper
strip. (All metals should be the factors in each step of the
same width.) If necessary, pol- procedure?
ish the metals with sandpaper
until they are shiny. 2. What were the results from

2. Using tongs, hold a paper clip each step of the procedure?
How do you interpret each
in the hottest part of the burn- result?
er flame for 30 s. Repeat with a
ball of steel wool 2 cm in
diameter.

SECTION REVIEW 4. What is meant by a rate law for a chemical
reaction? Explain the conditions under which
1. What is studied in the branch of chemistry that is a rate law can be written from a chemical
known as chemical kinetics? equation. When can a rate law not be written
from a single step?
2. List five important factors that influence the rate
of chemical reactions. Critical Thinking

3. What is a catalyst? Explain the effect of a catalyst 5. RELATING IDEAS Using the ideas of reaction ki-
on the rate of chemical reactions. How does a netics, explain the purpose of food refrigeration.
catalyst influence the activation energy required
by a particular reaction?

578 C H A P T E R 1 7