Biermann Ch. Handbook of Pulping and Papermaking

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324 13. INTRODUCTORY CHEMISTRY REVIEW

13.8 SOLUBILITY PRODUCTS

Many salts have limited solubility in water.

This leads to scale formation on processing equip-

ment. Conditions are chosen to minimize scale

formation by understanding the solubility proper-

ties of these compounds. For a salt of the form

A^B^,

the following relationship can be written:

AJ3^(s) ^ aA + bB. The effective concentration

of a solid in dilute aqueous solutions is a constant,

provided there is excess solid. The solubility

constant can be derived from the law of mass

action where the concentration (strictly speaking

activities) of the reactant (the solid, which is of

constant activity) is incorporated into the equilibri-

um constant and defined as follows:

Table 13-5. Solubility product constants at 25°C.

i^sp = [Ar[Br

(13-10)

If one of the ions is already in solution, the

solubility of

the

original solid will be much lower.

This is known as the common ion effect. For

example, the solubility of AgCl is much lower in

0.1 M HCl than it is in pure water. Eample 13

demonstrates this concept.

Table 13-5 gives some values of solubility

product constants. One should realize that differ-

ent sources give different values that may vary by

as much as a factor of

ten,

especially for extreme-

ly small values. Lange's Handbook of

Chemistry

lists many values and the CRC Handbook of

Chemistry and Physics gives calculated values of

^sp-

^sp values can be calculated from the rela-

tionship of In X,p = AG//?r, where AG is the

Gibbs free energy of the dissolution reaction.

This is the same relationship for any equilibrium

constant, of course.

EXAMPLE 13. BaS04 from wood causes much

scaling on processing equipment in the

southern

U.S.

^^p for barium sulfate is 1.1 x

10'^^

(mol/L)2 at 25

°C.

What is the concen-

tration of Ba^"^ and 804^" in distilled water

containing excess BaS04(s)?

SOLUTION: BaS04(s) ^ Ba^^ + SO,^'

[Ba^+l = [SO42-] =;c; jc' = 1.1 X

X = 1.05 X

10-5

10-^

Substance

Aluminum

1 hydroxide

Barium carbonate

Barium sulfate

1 Calcium carbonate

Calcium

hydroxide

Calcium sulfate

Calcium sulfite

Iron(II)

hydroxide

Iron(II) sulfide

Iron(III)

hydroxide

Magnesium

carbonate

Magnesium

hydroxide

Magnesium sulfite

Formula

A1(0H)3

BaCOj

BaS04

CaCOj

Ca(0H)2

CaS04

CaSOj

Fe(0H)2

FeS

Fe(0H)3

MgC03

Mg(0H)2

M^S03

^sp 1

1 X 10-33

10-^

1.1

10-^*^

10-^

5.5 X 10-^

10-5

7 X 10-«

1 X 10-^*

6.3

X 10-^«

3 X 10-3^

1 X 10-5

1.5 X 10-^*

3.2

10-3

PROBLEM: The concentration of sulfate from

sulfuric acid in Whitewater is

0.0001

What is the equilibrium concentration of

Ba^^

ion if excess BaS04 is present?

Answer: 1.1 x

10-^

M. This shows that the

presence of sulfate ion in solution reduces the

solubility of the barium sulfate, which is

predicted by the common-ion effect.

EXAMPLE 14. Calcium hydroxide is important

in the kraft process as it is used to regenerate

sodium hydroxide from sodium carbonate.

The K,^ for Ca(0H)2 is 5.5 x 10'^ (mol/L)^

at 25°C. What is the solubility of calcium

hydroxide in water?

SOLUTION: Ca(0H)2 ^ Ca^^ + 2 OH'; so

SOLUBILITY PRODUCTS 325

[Ca'-'ILOH-]^

= 5.5 X

10"^

[Ca2+] = x\ [OH] = 2x

X'{2xf

= 5.5 X

10-^

jc =

0.0111

M; [OH] =0.0222

the pH is 12.34 and the original concentra-

tion of OH' (10'^) does not interfere with its

solubilization by the common ion effect.

Salts containing anions that are basic (for

example, OH",

COj^",

or S^) have more complicat-

ed chemistry. For example ^sp for calcium car-

bonate is 4 X 10-^ (mol/L)^ at 25°C. However,

the carbonate anion is a weak base and at pH

10.32 half of the COa^" is actually in the form of

HCOj'.

Since calcium bicarbonate is fairly soluble

in water, below pH 7 calcium carbonate is fairly

soluble and exists as calciimi bicarbonate.

Salts may contain cations that are acidic

because they complex with OH" groups (leaving

H"^ behind). The solubilities of such salts may be

strongly dependent on pH. Aluminum will be

considered in detail because of its importance in

wet end chemistry. Aluminimi hydroxide has very

lunited solubility, as shown in Fig. 13-4, since:

A1(0H)3 (s) ^ AP-^ -h 3 OH"; ^sp = 1 x 10"^^

Since the solubility of the aluminum ion is of

concern in wet end chemistry, it will be deter-

mined as follows: [AP+] = ^,p x

[OH"]^

Since

K^ = 10"^* = [H-^][OH"], the molar solubility of

aluminum ion can be written as a function of pH

as follows: [AP+] = K,^ X ([H+]/X;,)^

Additionally, one can consider the solubility

of other aluminum species. At higher pH's the

formation of a higher complex occurs to an appre-

Fig. 13-4. Solubility of various simple aluminum ions as a function of pH.

326 13- INTRODUCTORY CHEMISTRY REVffiW

Table 13-6. Assigning oxidation numbers (ON) to atoms of molecules or ions.

No.

Rule

The total ON of all of the atoms in the molecule or ion is equal to the total charge of the

molecule or atom.

Atoms in their elemental form have an ON = 0

Group I elements in combined form have ON = +1

Group n elements in combined form have ON = +2

Group in elements (except boron) have ON = +1 or +3 depending on the charge of the

metal ion.

Group IV elements (except carbon and silicon) have ON = +4 or +2 depending on the

charge of the metal ion.

Hydrogen ON = +1 in combination with nonmetals and -1 in combination with metals.

Fluorine in combined form has an ON = -1

6. Oxygen has ON = -1 in peroxides

(Oj^',

for example, HOOH)

ON = -1/2 in superoxides (Oj)

ON = -1/3 in ozonides (O3)

ON = -2 unless combined with F

ciable extent. A1(0H)3 (s) + OH" ^ AKOH)/

(aq) with

= 40 at 25

°C.

The concentration of

aluminum is solved as Al(0H)4"(aq) = ^ x

{KJ\R^]),

At intermediate pH the aluminum ion

complexes according to AP+ + HjO -^ A1(0H)2+

+ H+ with K, = 9.78 x 10^; [A1(0H)2+] =

^a[Al^"']/[H''].

Other complexes (Section 14.7)

exist in the pH range of 4-6, but their concentra-

tions are too complicated to calculate. Fig. 13-4

shows the concentration of AP+, Al(OH)^"^, and

A1(0H)4" as a function of pH in terms of g/L

hydrated aluminum sulfate (alum, with a formula

weight of 666 grams per 2 moles of AP^.)

13.9 OXIDATION-REDUCTION REACTIONS

Oxidation-reduction (redox) reactions are

similar to acid-base reactions except the unit of

currency is not the proton but the electron. They

are important in combustion, redox titrations such

as,

determination of lignin contents using potassi-

um permanganate, and many other processes.

Like protons, electrons cannot exist in solution by

themselves so both electron donors and electron

acceptors are required for a reaction to occur.

Redox reactions, however, are not confined to

solvents such as water, but can occur in any form

of matter. Oxidation is defined as the loss of

electrons by an element during a chemical reac-

tion, and reduction is defined as the gain of

electrons. (This is easily remembered as LEO

growled GERRR; loss-electrons-oxidation, gain-

electrons-reduction.) The oxidizing agent, or

oxidant, is the substance that causes oxidation,

while the reducing agent is the substance that

causes reduction. Oxidation was originally a

reaction between a compound and oxygen gas,

where oxygen itself is reduced but causes oxida-

tion. Oxygen is the most electronegative element

(except for fluorine which is not nearly as ubiqui-

tous as oxygen). Disproportionation is a redox

reaction where a compound is both oxidized and

reduced, such as NO2 in the production of nitric

acid:

3NO2 + H2O -* 2HNO3 + NO

Electrons are assigned to atoms of molecules

or ions according to the rules in Table 13-6.

OXTOATION-REDUCTION REACTIONS 327

Oxidation numbers (ON) are assigned to elements

with these rules. The lower number rule always

takes precedence over a later numbered

rule.

Rule

6 is sunmiarized as follows: except for hydrogen

peroxide where oxygen has a charge of-1, oxygen

ahnost always has a charge of -2. While these

rules are somewhat arbitrary, it allows reactions

involving transfer of electrons to be balanced. Let

us consider the reaction of sodium metal with

water used to make sodium hydroxide. 2Na +

2H2O -* 2Na"^ + 20H- + Hj. Clearly sodium has

gone from neutral to a

+ charge by losing an

electron; in this reaction sodium was the reducing

agent (for hydrogen) and was oxidized. Some of

the hydrogen of water went from

+ to neutral by

accepting an electron; hydrogen was the oxidizing

agent, and it was reduced in the reaction.

EXAMPLE 15. Calculate the charge on sulfur in

each of the following compounds: Sg, HjS,

SO2,

H2SO3, and SO3.

SOLUTION: In each case hydrogen has an ON

= 4-1 and oxygen = -2; therefore the charg-

es on sulfur are 0, -2, +4, +4, +6. The

chemistry of sulfur is relatively complex due

to all of the possible oxidation numbers in a

variety of compounds.

PROBLEM: What are the oxidation numbers of

carbon in each of the following compounds:

C, CO, CO2, and H2CO3?

Answer: 0, +2, +4, +4.

One can determine whether or not a redox

reaction has occurred by determining whether the

oxidation number of any element has changed in

the course of a reaction. For example, the reac-

tion of

H"*"

-f OH" -* H2O is not a redox reaction,

but C + O2

-»•

CO2 is a redox reaction.

13.10 ELECTROCHEMISTRY

Many materials are capable of both oxidation

and reduction reactions. The type of reaction de-

pends on what the other reactant

is.

For example,

sulfur may be reduced with hydrogen to form HjS

or oxidized with oxygen to form SOj. Oxygen,

being more electronegative than sulfur, causes

oxidation, while hydrogen is more electropositive

than sulfur and causes reduction. Therefore,

redox reactions are conveniently split into two

half

reactions with one half reaction corresponding to

oxidation and the other half reaction corresponding

to reduction. Zinc metal dissolves in hydrochloric

acid with the overall reaction as follows:

Zn(s) + 2 HCl ^ ZnCl2 + H2(g) (overall)

Zn(s) -> Zrf+ + 2e- (oxidation)

2H+ + 2e- -> H,

(reduction)

It is useful to prepare a table of half reactions

with the energy released or absorbed in the reac-

tion to predict whether or not overall reactions will

occur based on thermodynamics. This technique

also simplifies the balancing of complex chemical

equations. Lange

Handbook of

Chemistry

is one

good source for half reaction potentials. Some

half reactions are listed in Table 13-7 with com-

pounds on the left in decreasing order of oxidation

power. This means the higher a compound is on

the left hand portion of the table, the more likely

it is to induce oxidation of another compound or

ion. Similarly, compounds or ions on the lower

right are powerful reducing agents, the compounds

higher up on the right hand side of the table are

lower in reduction potential. Remember these are

thermodynamic values; they say nothing about the

reaction kinetics.

EXAMPLE 16. Balance the following equation,

which represents the Palmrose method of

titrating sulfite liquors:

H2SO3 + KIO3 ^ H2SO4 + KI

SOLUTION: First write the two half reactions.

Each mole of sulfur in sulfurous acid looses

two electrons as it goes from +4 to +6 va-

lence. Each mole of iodine gains six elec-

trons as it goes from +5 to

-1.

So,

KIO3 + 6e- + 6H+ -^ 3H2O + KI

3H2SO3 + 3H20-> 3H2SO4 + 6H+ +6e-

3H2SO3 +

KI03-^

3H2SO4 + KI

328 13. INTRODUCTORY CHEMISTRY REVIEW

Table 13-7. Reduction half reaction potentials at 25^C.

Oxidizing agent

1 Strongly oxidizing

HCIO

+ H+

Mn04-

+ 8H+

HCIO

+ H+

CIj

CIO2

+ H+

lOj-

+ 6H+

IO3-

+ 6H-^

CIO2

CIO

+ H2O

Fe'^

ClOj-

+ 4H+

IjCaq)

2H2O

Cu^+

S04^-

+ 4H+

SA^-

an-'

Fe'^

Ni^^

Fe^+

Cr3+

Zrf+

2H2O

Mrf+

2SO3'-

+ 2H2O

AP+

Mg^+

Na-*-

Ca^+

K+

+ e-»

+ 5e-^

+ 2e--»

+ 2e-^

+ e--»

+ 5e-^

+ 6e-^

+ e-^

+ 2e--*

+ e-^

+ 4e--*

+ 2e--»

+ 4e--»

+ 2e--*

+ 2e-^

+ 2e--»

+ 2e"-

+ 3e-^

+ 2e-^

+ 2e--»

+ 2e-^

+ 3e-^

2er-*

+ 2e-^

+ 2e--»

+ 2e-^

+ 3e--*

+ 2e--»

+ e-^

+ 2e-^

+ e-^

Reducing agent

1/2

CI2

+ H2O

Mrf+ + 2H2O

CI-

+ H2O

2C1-

HCIO2

I2 +

3H2O

I-

3H2O

HCA"

CI-

+ 2 OH-

2 OH

Fe^+

CI + 2H2O

21-

3 1-

4 0H-

SO2 + 2H2O (or H2SO3)

28203^-

(thiosulfate, hypo)

S2^-

+ 2

OH-

S2O4'-

+ 40H-

(hydrosulfite)

^Strongly reducing

E",

V at 25°

+ 1.63

+ 1.51

+ 1.49

+ 1.36

+ 1.27

+ 1.19

sum of two reactions

+0.95

+0.89

+0.88

+0.77

+0.75

+0.620

+0.534

+0.41(liVNaOH)

+0.345

+0.17

+0.08

0.00 (by definition)

-0.04

-0.250

-0.43

-0.440

-0.71

-0.762

-0.83

-1.05

-1.12

-1.66

-2.36

-2.712

-2.87

-2.922

ELECTROCHEMISTRY 329

ANODE

ELECTRONS

<

CATIONS

ANIONS -^

SALT BRIDGE

OXIDATION

CATHODE

REDUCTION

Fig. 13-5. Diagram of the Daniell cell, a repre-

sentative electrochemical cell.

Two half reactions can form a redox couple

which produces a voltage. By convention, the

oxidation cell is written first. The classic electro-

chemical cell is the Daniell cell invented in 1836

to provide power for the telegraph industry. It is

a zinc-copper cell. A representative cell is con-

structed as shown in Fig. 13-5.

Cell diagrams are written with the an-

ode

cathode half reactions. The chemical equa-

tion and cell diagram of the Daniell cell are:

Zn + Ctf-^ <:i Cu + Zrf+

Zn|Zrf-^(aq)||Cu2+(aq)|Cu

If a solid metal is not an electrode for a

particular half reaction, platinum should be speci-

fied to provide the necessary electrical contact.

For example, the reaction of zinc metal with

hydrochloric acid is specified as:

Zn|Zrf^(aq)||H^(aq)|H2(g)|Pt

E° is the standard

cell

potential, or voltage,

generated when the activity of all of the species is

equal to unity and the pressure is 1 atm, i.e., the

concentrations of all (non gas) species are essen-

tially 1 M. The cathode is the electrode where

reduction occurs, and the anode is the electrode

where oxidation occurs. The anode is the source

of electrons and is the negative terminal.

£°(anode) is -£^° of the reduction reaction.

£:°(cell) = £°(anode) + £:°(cathode) (13-11)

EXAMPLE 17. What is

E""

of the voltaic cell

Zn|Zrf+(l A^fFe^-'d A^|Fe?

SOLUTION:

3(Zn ^ Zrf+ + If)

2{Ve^ + 3e- -* Fe)

+0.76

-0.04

+0.72

PROBLEM: What voltage is generated by a

lithium cell operating under standard condi-

tions with the reaction:

2Li + CI2 ^ 2Li+ + 2C1-

Answer: 4.4 V.

The equilibrium constant for any reaction

which can be expressed as two half reactions of

known potential can be derived by substituting AG

-nFE""

into the equation AG= -RT In K to

give:

InK^nF/RTxE''

(13-12)

where F is the number of coulombs in a mole of

electrons or 96,485

C/mol

and n is the number of

electrons transferred in the reaction. Since 1 J =

1 C X 1 V, the term nFE° is the energy of the

reaction in joules. The term F/RT appears fre-

quently in electrochemistry and is equal to 0.02569

V at 25°C. Eq. 13-12 is usually written as a

common logarithm and takes on the form:

log K = nEVO.0592 V 25°C (13-13)

The Nernst

equation

relates the actual voltage

of the cell to the standard voltage of the cell and

the concentration of reactants and products. The

reaction quotient, Q, is the actual concentration of

products raised to the stoichiometric coefficients

divided by the concentration of reactants raised to

the stoichiometric coefficients in the usual manner.

^ = ^(ceU)

0.0592 V

(13-13a)

xlog<?

330

13.

INTRODUCTORY CHEMISTRY REVIEW

EXAMPLE 18. Give the equilibrium constant

for the reaction: Ikg" + Zn -* Zrf+ + 2Ag

SOLUTION: E'' = 1.56 V;

log K = 2(1.56)/0.0592 = 52.7;

EXAMPLE 19. What is the potential of a Zn-

Cu^^ cell with [Cu^+J = 2 M and [Zrf+] =

0.2 M?

SOLUTION: £ = 1.10 - 0.0592/2 x log 0.2/2

= 1.13 V.

13.11 PRACTICAL ASPECTS OF

ELECTROCHEMISTRY

Corrosion

Corrosion of metals is the result of electro-

chemical reactions. For example, in the zinc-

copper cell, the zinc metal is lost or corroded.

Since metal becomes soluble (corroded) as it goes

from a neutral to cationic state, corrosion occurs

at the anode where electrons are given up by

metal. These electrons must be used by a cathodic

reaction to complete the cell. An obvious corro-

sion reaction occurs when zinc or iron is put into

hydrochloric acid. Numerous bubbles form as the

metal displaces hydrogen from water. Parts of the

metal act as the anode where electrons are given

up and the metal goes into solution as ions, while

other parts of the metal act as the cathode where

electrons are accepted to convert hydrogen ions

into hydrogen gas. As the anode and cathode

areas move around the surface, the entire piece of

metal is ultimately dissolved.

In water near neutral pH, hydroxide ion is

produced at the cathode (by loss of hydrogen ions)

which precipitates ferrous ion in solution to give

ferrous hydroxide. The ferrous hydroxide is

further oxidized to ferric hydroxide, which is the

rust observed on the surface of iron. The behav-

ior of die rust (whether it forms on the surface or

away from the surface) depends on the pH and

oxygen content of the water and gives rise to

several types of corrosion mentioned below.

When two metals of different composition

(dissunilar metals) are electrically connected and

placed in a solution with an electrolyte, a voltage

is generated. The resultant corrosion is called

galvanic action. This can occur as potentials exist

across various parts of

the

same piece of metal due

to differences in surface compositions (like varia-

tions in oxygen concentrations caused by dirt that

excludes oxygen, leading to pitting), solution

composition at the surface, stress in the metal, etc.

When two dissimilar metals are in contact, the

more strongly reducing metal (Table 13-7) will

corrode. A large difference in potential may lead

to severe corrosion. For example, zinc will

corrode if attached to iron. When the surface area

of the metal that corrodes is small relative to the

other metal, corrosion is particularly quick. For

example, iron rivets holding copper sheets together

will corrode very quickly. Also, pitting represents

a rapid corrosion of a small area made anionic by

oxygen starvation. (These are thermodynamic

considerations, they do not say anything about the

rate of corrosion.)

Often an easily corroded metal is deliberately

attached to a large piece of metal to protect the

large piece of metal from corroding. The metal

that is corroded is called a sacrifice anode and

keeps the metal it protects cathodic, thereby not

allowing it to corrode. This technique is known as

cathodic protection. Galvanized (zinc coated)

iron, magnesium blocks inside hot water tanks,

and zinc anodes attached to the submerged metal

parts of marine vessels are three examples of

sacrifice anodes. It is used in water heaters,

marine vessels and

engines,

pipelines, and bridges.

The sacrificial anode (usually magnesium or zinc)

corrodes in preference to the material that it

protects. The sacrificial anode is periodically

replaced to maintain protection.

A related method of protection is to apply a

small, negative direct current voltage to the device

being protected if it is electrically insulated from

its environment. It must be insulated from its

environment or else it would take too much power

(amperage) to maintain the voltage. The applied

voltage makes the metal surface cathodic. This

technique has been used to protect underground

metal pipes (if they are electrically insulated from

ground).

PRACTICAL ASPECTS OF ELECTROCHEMISTRY 331

A more strongly reducing metal can displace

another metal from solution. A copper film forms

over most of a piece of iron that is dipped in

copper sulfate. If a surface of a less strongly

reducing metal is desired, the reaction can be

forced by the application of an external voltage in

the process of electroplating. Electroplating

occurs when reduction of a metal from solution is

forced by an external voltage, much like the

reaction at the cathode in a Daniell cell.

Some materials, like aluminum, would seem

to be unstable based on their position in the elec-

trochemical series. However, aluminum and chro-

mium both form oxides on their surfaces that are

very hard, insoluble, nonporous, self-healing, and

inert. The formation of such films renders metals

passive. These surfaces protect the underlying

metal from corrosion. For example, 12-18%

chromium in stainless steels protects the accompa-

nying iron from corrosion. Actually, stainless

steel (active) is vulnerable to corrosion; however,

the coating (passive) that forms on stainless steel

is inert. Crevice corrosion results when oxygen is

kept from the surface of an active material. Oxy-

gen gains electrons during its reaction so high

areas of oxygen react at the cathode and low areas

of oxygen provide an area of decreased potential

for the anode reactions. This occurs under depos-

its,

barnacles, or plastic washers used to insulate

fastenings on stainless steel plates used in sea

water. Table 13-8 shows metals presented in

increasing resistance to galvanic corrosion in sea

water. In other solutions the order may change.

Corrosion can be decreased in a number of

ways.

Painting prevents air, water, and salts from

reaching the surface of the metal. Galvanization

is the process of coating the entire surface with an

unbroken coat of

zinc

by immersion in molten zinc

or by electroplating. The zinc is protected by

passivation with a zinc oxide coating. The iron is

protected since zinc is oxidized in preference to it.

Electrolysis

Some other practical aspects of electrochem-

istry include production of NaOH and Clj and pro-

duction of

ClOj.

Ion-selective electrodes

Ion-selective electrodes are cells that obey the

Table 13-8. Galvanic series of metals and

alloys in sea water with increasing resistance.

Metal or alloy

Magnesium

Zinc

Aluminum alloys

Aluminum

Cadmium

Mild Steel

Alloy Steel

Cast Iron

Type 410 SS, Active

Type 430 SS, Active

Type 304 SS, Active

Type 316 SS, Active

Nickel Cast Iron

Yellow Brass

Copper

Nickel

Type 304 SS, passive

Type 316 SS, passive

Monel

Silver

Hastelloy, Alloy C

Titaniimi

Platinum

Nominal composition

Low carbon

Various below 5%

3-4% C, 0.5-3.5% Si

11.5-13.5%

12-16%

18%

Cr, 8% Ni

18%Cr,8%Ni,3%Mo

0.5-10% Ni, 3-4% C,Cr

65%

Cu, 35% Zn

18%

Cr, 8% Ni

18%Cr,8%Ni,3%Mo

67%

Ni, 28% Cu

Ni-Cr-Mo

rules of electrochemistry. The Nemst equation

shows that the electrochemical potential of a cell

depends on the chemical concentration of the

reactive species. This principle is used in pH

electrodes and other ion-specific electrodes for

sodium, sulfide, carbonate, fluoride, etc. Ion-

selective electrodes, other than for pH, have not

been used extensively by the pulp and paper indus-

try, but they offer the potential for rapid laborato-

ry analysis or on-line sensors for process control.

13.12 PROPERTIES OF GASES

Ideal gas law

The kinetic theory of gases is a set of three

assumptions about the behavior of ideal gases.

While the assumptions are not completely true.

332 13- INTRODUCTORY CHEMISTRY REVIEW

they are approximately true at low pressures,

generally below one atmosphere pressure. (Much

the

non-ideal behavior of

gases

is a result of

the

nonzero volume of

molecules,

which increases the

pressure compared with ideal behavior, and the

formation of dimers, trimers, and clusters, which

decrease the actual pressure compared with ideal

behavior.) From these assumptions, much about

gas behavior can be predicted. The assumptions

are:

A gas consists of a collection of mole-

cules or atoms in continuous random motion.

The molecules or atoms are infinitely

small particles which move in straight lines

until they collide.

Gas molecules do not affect each other

except during collisions.

The ideal gas law relates the temperature,

pressure, volume,

and

nimiber of moles of

ideal

gas.

always expressed in the absolute temper-

ature scale of Kelvin, while volume is usually

expressed in liters. When pressure is in atmo-

spheres, R = 0.08206 L-atm-K-^-mor^; when

pressure is in torr, R = 62.37 L-torr-K'^-mol'^;

when pressure is in Pascals, R = 8.314

J-K"^-

mol'^ R is called the gas

constant.

The ideal gas

law is usually written as Eq. 13-14. Standard

temperature and pressure, STP, conditions are

defined as 273.15 K and 760 torr (or 1 atm);

under these conditions one mole of an ideal gas

occupies 22.41 L.

EXAMPLE

20.

1.5 g of argon occupies 4.03 L at

298 K. What is the pressure exerted by this

gas inside the vessel, in atm, if argon be-

haves as an ideal gas?

SOLUTION: First the number of moles of argon

is calculated to be 0.0375 based on its atomic

weight of 40. Using Eq. 13-14 with R =

0.08206 L-atm'K"^-mol

the pressure is de-

termined to be

0.228

atm.

PROBLEM: What is the volume of 0.154 mol of

an ideal gas at 745 torr and 30.0°C? An-

swer: 3.91 L.

Often one desires to compare a single system

under two different states of pressure, volume, and

temperature. In this case the two states will be

designated as

and 2 with subscripts. Rearrange-

ment of Eq. 13-14, with nR a constant, leads to

Eq.

13-16.

PiVi/r, =

P2V2IT2

(13-16)

EXAMPLE 21. 5.0 liters of a gas are collected

at a temperature of 300 K and a pressure of

720 torr. What would the volume be at STP

conditions?

SOLUTION:

PV= nRT

(13-14)

.^^^720tMr^273K

^760 torr'^300K

=4.315

The compressibility factor, Z, is defined in Eq.

13-15,

and is equal to 1 for an ideal gas. Z is a

measure of the deviation from ideal behavior and

is usually less than one except for highly volatile

gases at low pressures.

Z = PV/nRT

(13-15)

The root-mean-square velocity of the mole-

cules V is related to the molar mass M by the

relationship v =

(3RTIM)^^.

Since diffusion is

proportional to velocity, it is also inversely pro-

portional to the square root of molecular weight

(Graham's law). Since kinetic energy is Vimv^,

PROPERTIES OF GASES 333

it follows that the internal energy (not to be con-

fused with work done by an expanding gas which

is an external energy) of a gas is

3/2'RT.

The

latter equation can be used to calculate the heat

capacity of an ideal gas.

Real gas equations

There are many empirical equations that

approximate the behavior of real gases at elevated

pressures where deviations from ideal become

appreciable. The most famous is the van der

Waals equation developed by the nineteenth

century Dutch chemist. The constant a represents

the attractive forces which decrease the actual

pressure from ideal, and the constant b represents

the effect of repulsions or the finite volume of the

molecules which increase the actual pressure from

ideal. The equation is not useful for the liquid

phase.

The equation is not useful at very high pres-

sures;

for example, while the van der Waals equa-

tion predicts a compressibility factor, Z, of 0.375

at the critical point of a gas, in fact, gases usually

have a compressibility factor on the order of 0.25

to 0.30 at their critical point. The

CRC

Handbook

Chemistry

and Physics gives values of

and 6

for many gases (Table 13-9). The constants may

also be obtained from the critical data of gases,

usually using P^ and

T^.

as follows:

a «

21R^T^I6AP,\

b « RTJ^P,

One of

the

most accurate equations of state is

the

Redlich'Kwong

equation,

which is useful over

wide ranges of temperature and pressure. The

constants a and b are different than those of the

van der Waals equation.

Table 13-9. Van der

Waals'

constants of sever-

al gases.

(P +

V(V+bln)T^^

)

(Y-nb)

= nRT

13.13 ANNOTATED BIBLIOGRAPHY

Corrosion

Perry's

Chemical Engineers * Handbook

has

Gas

1 COj

CI2

loj

1 (^113)20^2

SO2

H2O

CH4

a L^-atm-mol"^

3.59

6.49

1.39

1.36

8.66

6.71

5.46

2.25

L-mol' 1

0.0427

0.0562

0.0391

0.0318

0.0845

0.0564

0.0305

0.0428

a very useful, detailed chapter on the subject

(section 23 of the 6th ed.).

Corrosion in Action, International Nickel

Co.,

1961. This was used as a source of

information about corrosion for this chapter.

EXERCISES

The

elements

Draw the Lewis structures for the following

compounds showing all valence electrons.

a) NajS

c) CCI4

b) CO2

d) CH4

Ionic and

covalent bonds

Which of the following statements is true of

a solution containing an electrolyte?

a) An electrolyte solution contains ions.

b) Water is a good solvent to use to pro-

duce an electrolyte solution.

c) An electrolyte solution will conduct

electricity.

d) All of the above.

Which of the following is the most ionic

compound?

a) BF3

c) Na2S

b) CCI4

d) Si02