STANDARD ATOMIC MASSES 1979(Scaled to the relative atomic mass , A ,.(I2C) = 12)Name Actinium Aluminium Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Boron Bromine Cadmium Caesium Calcium Californium Carbon Cerium Chlorine Chromium Cobalt Copper Curium Dysprosium Einsteinium Erbium Europium Fermium Fluorine Francium Gadolinium Gallium Germanium Gold Hafnium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium Lead Lithium Lutetium Magnesium Manganese Mendelevium Mercury Atomic Symbol number Ac 89 Al 13 Am 95 Sb 51 Ar 18 As 33 At 85 Ba 56 Bk 97 Be 4 Bi 83 B 5 Br 35 Cd 48 55 Cs Ca 20 Cf 98 C 6 Ce 58 Cl 17 Cr 24 Co 27 Cu 29 Cm 96 Dy 66 Es 99 Er 68 Eu 63 Fm 100 F 9 Fr 87 Gd 64 Ga 31 Ge 32 Au 79 Hf 72 He 2 Ho 67 H I In 49 I 53 Ir 77 Fe 26 Kr 36 La 57 Lr 103 Pb 82 Li 3 Lu 71 Mg 12 Mn 25 Md 101 Hg 80 Atomic mass 227.0278 26 .98154 (243) 121.75* 39 .948 74.9216 (210) 137 .33 (247) 9.01218 208.9804 10.81 79 .904 112.41 132 .9054 40 .08 (25 I) 12.011 140. 12 35.453 51.996 58 .9332 63 .546* (247) 162 .50* (252) 167.26* 151.96 (257) 18.998403 (223) 157.25 * 69.72 72 .59* 196.9665 178.49* 4.00260 164.9304 1.0079 114.82 126.9045 192 .22 * 55 .847* 83 .80 138.9055 * (260) 207 .2 6.941 * 174 .967 * 24.305 54.9380 (258) 200 .59* Name Molybdenum Neodymium Neon Neptunium Nickel Niobium Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum Plutonium Polonium Potassium Praseodymium Promethium Protactinium Radium Radon Rhenium Rhodium Rubidium Ruthenium Samarium Scandium Selenium Silicon Silver Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten (U nnilhexium) (Unnilpentium) (U nnilquadium) Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium Atomic Symbol number Mo Nd Ne Np Ni Nb N No Os 42 60 10 93 28 41 7 102 76 8 46 15 78 94 84 19 59 61 91 88 86 75 45 37 44 62 21 34 14 47 II 38 16 73 43 52 65 81 90 69 50 22 74 106 105 104 92 23 54 70 39 30 40 Atomic mass 95.94 144.24* 20 . 179 237 .0482 58.69 92.9064 14 .0067 (259) 190.2 15 .9994* 106.42 30.97376 195.08* (244) (209) 39.0983 140.9077 (145) 231 .0359 226 .0254 (222) 186.207 102 .9055 85.4678 * 101.07* 150.36* 44.9559 78 .96* 28.0855 * 107 .868 22 .98977 87. 62 32 .06 180.9479 (98) 127.60* 158.9254 204 .383 232 .0381 168.9342 118 .69* 47 .88 * 183.85* (263) (262) (261) 238.0289 50.9415 131.29* 173 .04 * 88 .9059 65.38 91.22

0Pd P Pt Pu Po K Pr Pm Pa Ra Rn Re Rh Rb Ru Sm Sc Se Si Ag Na Sr S Ta Tc Te Tb TI Th Tm Sn Ti

W(Unh) (Unp) (Unq) U V Xe Yb Y Zn Zr

Source: Pure and Applied Chemistry , 51, 405 (1979 ). By permission . Value s are considered reliable to I in the last digit or 3 when followed by an asterisk(*). Values in parentheses are used for radioactive elements whose atomic weight s cannot be quoted precisel y without knowledge of the origin of the elements; the value given is the atomic mass number of the isotope of th at element of longest known half-life.

FUNDAMENTAL CONSTANTS(approximate values; best values are in Appendix IV) Quantity Gas constant Zero of the Celsius scale Standard atmosphere Standard molar volume of ideal gas A vogadro constant Boltzmann constant Standard acceleration of gravity Elementary charge Faraday constant Speed of light in vacuum Planck constant Rest mass of electron Permittivity of vacuum SymbolR

Value 8.314 J K- 1 mol-I 273.15 K 1.013 x 105 Pa 22.41 x 10- 3 m3 mol-I 6.022 x 1023 mol I 1.381 x 10- 23 J K- 1 9.807 m s -2

ToPo

Vo

=

RTolpo

!

,

e

F c Il IiIn

=

NAe

=

h121T'

en

Bohr radius Hartree energy"-

41T'eo 1/41T'eo ao = 41T'eoIi2/me2Eh = el l41T'e oao

1.602 9.648 2.998 6.626 1.055 9.110 8.854 LIB 8.988 5.292 4.360

'>\.

X

x xX

X'>\.

X

x xx:

10 19 C 104 C mol-I lOR m s I 10 34 J s 10- 34 J s 10- 31 kg 10- 12 C2 N- 1 m 2 10- 10 C 2 N- I m -2 109 N m 2 C- 2 10 II m 10 I~ J

CONVERSION FACTORS1 L = 10- 3 m' (exactly) = 1 dm 3 I atm = 1.01325 Pa (exactly) I atm = 760 Torr (exactly) 1 Torr = 1.000 mmHg 1 cal = 4.184 J (exactly) 1 erg = 1 dyne cm = 10- 7 J (exactly) 1 eV = 96.48456 kJ/mol1 A = 10 10 m = 0.1 nm = 100 pm I inch = 2.54 cm (exactly) 1 pound = 453.6 g I gallon = 3.785 L 1 Btu = 1.055 kJ I hp = 746 W

i

I

MATHEMATICAL DATA1T =

3.14159265 ...

e = 2.7182818 ...(all x)

In x = 2.302585 ... log x

In I(l

+

x)

=x

-

Y~x"

+X~

IA\"3 -

Y4X 4

+(x~

(x"

< 1)

+

x )-1x )-1

- X

(l (I -

+

X

+ +

x" -

X3 X3

+ +4x 3

>

In the kinetic theory of gases we deal with integrals of the general type

(4. 3 5)

(4. 36)

M athematica l I nterl u d e

63

so that

In C[3) =

Loo x2n+ 1 e-Px2

dx =

n !)[3 - (n + 1 ).

(4. 37)

The higher-order integrals can be obtained from those of lower order by differentiation ; differentiating Eq. with respect to [3 yields

(4. 3 7)

dU[3) d[3

or

(4. 3 8)Two cases commonly arise. In this case we apply Eq.

(4. 3 7) directly and no difficulty ensues. The lowest member is 1p . All other members can be obtained from Eq. (4. 3 7) or by differentiating and using Eq. (4. 3 8).1 0 ([3) = -1 1 0 ([3)Case n.n

Case I.

n

= or a positive integer.

=

In this case we may also use Eq. directly, but unless we know the value ofthe factorial function for half-integral values of the argument we will be in trouble. If = - 1, the function takes the form

- 1, 1, t or n =

m

(4. 3 7)

- 1 where m = or a positive integer.

When

m 0, we have

=

Im - 1 /2 ([3) =

Loo x2me- Px2

n m

dx =

H(m - 1) !][3 - (m+ 1 /2).

(4. 39)( .

where in the second writing, x = [3 - 1 / 2 y, has been used. Comparing this result with the last member of Eq. we find that

L 1 /2 ([3) =

Loo e-px2 d

1 x = [3 - /2

LOO e - y2

dy = [3 - 1 / 2 L 1 /z (1),

4 40

)

(4. 39)

L 1 /2 (1) =

Loo e- y2and

dy = 1(

- 1)! .

(4. 4 1)

The integral, 1 - 1 / 2 (1), cannot be evaluated by elementary methods. We proceed by writing the integral in two ways,1 _ 1/ 2 (1) =

LOO e- x2

dx

then multiply them together to obtain,1 1 /2 (1) =

The integration is over the area of the first quadrant; we change variables to r 2

Loo LXle- (x2 + y2)

dx dy.

= x 2 + y2

64

T h e Struct u re of G ases

0 I1 /2 (1 ) = f/2d IXl e - r2r dr = G) 1''' e - r2 d(r2) = IXle - Y dy. The last integral is equal to O ! = 1; taking the square root of both sides, we have (4.42) I - l li l) = !In. Comparing Eqs. (4. 4 1) and (4.42), it follows that ( !) ! = In ; now from Eqs. (4. 40) and (4.42), L1 /2 (P) = !Jn p - 1 /2 . By differentiation, and by using Eq. (4. 3 8) we obtain dL 1 / n I 1 /2(P) - - --2 - 2yC (1.2 p - 3/2) dP_

and replace dx dy by the element of area in polar coordinates, r d dr. To cover the first quadrant we integrate from zero to and r from to 00 : the integral becomes

nl2

1.

1 I3/2(P) = dIdP/2 = 2yfir (1.. 1.2 p - 5/2). 2 Repetition of this procedure ultimately yields (P) = f oo x2me - PX2 dx = 2yfir 2(2m) .! p - (m + 1 /2) (4.43) Im - 1 /2 2m By comparing this result with Eq. (4. 3 9) we obtain the interesting result for half-integral factorials ( 2) = y n 2(2m) !, . (4.44) 2m Table 4.1 collects the most commonly used formulas._

and

1.

n

o

1.

n

1 m.

.

m

_

1. 1

C

m.

Ta b l e 4 . 1 i nteg ra ls that occ u r i n t h e k i netic t h e o ry of gases

(6) (7) (3) (4) (5)

1''' 1''' foOO

x 2 e - px2 dx = tfi tP - 3 / 2 x4 e - px2 dx = tfiiP - S / 2 (2n) !p - (n + l / 2 ) 2 2 nn !-

(8) (9)

x 2 ne - Px2 dx = 1.fi 2

f oo fooo 1 1-

x 2 n + l e - px2 dx = O xe - px2 dx = tP - 1

00

O

x 3 e - px 2 dx = tP - 2 x S e - px2 dx = p - 3

00

M athematica l I nterl u d e

65

* 4 . 7 . 1 The E r r o r F u n ct i o n

W e frequently have occasion t o use integrals o f the type o f Case I I above in which the upper limit is not extended to infinity but only to some finite value. These integrals are related to the error function (erf). We define erf (x) If the upper limit is extended to x --+=

00,

the integral is t.fi so that=

fe-U2 duo1.

(4.45)

erf (oo)

Thus a s x varies from zero t o infinity, erf (x) varies from zero t o unity. Ifwe add the integral from x to 00 multiplied by 2/.fi to both sides of the equation, we obtain erf (x) + Therefore

.fi foo e- u 2 du "2

=

[f0"e- u2 du foo e-U2 du] .fi "2+

=

2 .fi

fooe- u2 du 0

=

1.

2

.fi

fooe- u2 du"=

=

1

-

erf (x).

We define the co-error function, erfc (x), by erfc (x) Thus 1-

erf (x).

(4.46)

(4.47) Some values of the error function are given in Table 4.2.

Ta b l e 4 . 2 The error fu nct i o n :erf(x)=

In 0 e - ,,2 du2

I

X

x0.00 0. 10 0.20 0.30 0.40 0.50 0.60 0.70

erf(x)0.000 0. 1 12 0.223 0.329 0.428 0.521 0.604 0.678

x0.80 0.90 1.00 1 . 10 1 .20 1.30 1.40 1.50

erf(x)0.742 0.797 0.843 0.880 0.9 10 0.934 0.952 0.966

x1.60 1.70 1 .80 1 .90 2.00 2.20 2.40 2.50

erf(x)0.976 0.984 0.989 0.993 0.995 0.998 0.9993 0.9996

66

The Structu re of G asesE VA L U AT I O N O F A A N D

4.8

The constants and are determined by requiring that the distribution yield correct values of the total number of molecules and the average kinetic energy. The total number of molecules is obtained by summing over all possible values of between zero and infinity :

A {3

P

(4.48) The average kinetic energy is calculated by multiplying the kinetic energy, tmc 2 , by the number of molecules, dnc , which have that kinetic energy, summing over all values of c, and dividing by N, the total number of molecules. =OO dn fc:.C. =-=o tmc2__c (4.49) < f) _ - ----N-::--= Equations (4. 4 8) and (4. 4 9) determine A and {3. Replacing dnc in Eq. (4. 4 8) by the value given by Eq. (4. 3 4), we have N Loo4nNA 3 e - fJc2c2 dc. Dividing through by N and removing constants from under the integral sign yields 1 4nA 3 Loo c2 e - fJC2 dc. From Table 4.1 we have s: c 2 e - fJc 2 dc n l / 2 /4{3 3 / 2 . Hence, 1 4nA 3 n l / 2 /4{3 3 / 2 . So finally (4. 50) which gives the value of A 3 in terms of {3. In the second condition, Eq. (4. 4 9), we use the value for dnc from Eq. (4. 3 4) : foOO tmc24nNA 3 e -fJc2c2 dc ( 8.38 ) o. T2 T4 T3 The sum on the left-hand side of Eq. ( 8.37) consists of a number of terms, some positive and some negative. But the positive ones just balance the negative ones, and the sum is zero. We have to find numbers (temperatures) such that by dividing each term in Eq. (8.37) by a proper number we can obtain a sum in which the positive terms predominate, and thus fulfill the requirement of the inequality (8.38). We can make the positive terms predominate if we divide the positive terms in Eq. ( 8.37) by small numbers and the negative terms by larger numbers. However, this means that we are associating positive values of Q with low temperatures and negative values with high temperatures. This implies that heat is extracted from reservoirs at low temperatures and rejected to reservoirs at higher temperatures in the operation of the composite engine. The composite engine is conse quently an impossible engine, and our assumption, Eq. (8.33), must be incorrect. It follows that for any engine E ' , Ql Tl and

(8.37)+

We distinguish two cases :Case I.

l dQ' 'I T o .

( 8.39)

The engine E' is reversible. We have excluded the possibility expressed by ( 8.33 ) . If we assume that for E ' dQ ' - < 0' T

f

then we can reverse this engine, which changes all the signs but not the magnitudes of the Q ' s. Then we have dQ ' - > 0' T

f

and the proof is the same as before. This forces us to the conclusion that for any system

Therefore every system has a state property S, the entropy, such that

f dev = 0

(all reversible cycles).

( 8 . 40)

dS = dQ rev .T The study of the properties of the entropy will be undertaken in the next chapter.

( 8.41 )

The C l a us i u s I n eq u a l ity

1 67

Case II.

The engine E' is not reversible. For any engine we have only the possibilities expressed by (8.39). We have shown that the equality holds for the reversible engine. Since the heat and work effects associated with an irreversible cycle are different from those associated with a reversible cycle, this implies that the value of f for an irreversible cycle is different from the value, zero, associated with the reversible cycle. We have shown that for any engine the value cannot be greater than zero ; consequently, it must be less than zero. Therefore for irreversible cycles we must have

IlQ/T

(all irreversible cycles).8 . 1 5 T H E C LA U S I U S I N E Q U A L I TY

( 8.42)

Consider the following cycle : A system is transformed irreversibly from state 1 to state 2, then restored reversibly from state 2 to state 1. The cyclic integral is

J J1 and it is less than zero, by (8.42), since it is an irreversible cycle. Using the definition of dS, this relation becomes

l llTQ = (2 IlQTirr I I IlQTrev 2+ + pp. The high-pressureregion expands at the expense of the low-pressure region. The equilibrium requirement

T h e P ropert i es of G

21 3

is that

dA = 0 ; that is,

This is the condition for mechanical equilibrium ; namely, that the pressure have the same value in all parts of the system.1 0 . 8 T H E P R O P E RT I E S O F G

Pa = pp.

The fundamental equation (10.22),

views the Gibbs energy as a function of temperature and pressure ; the equivalent expres sion is therefore

dG =

-

S dT + V dp,+

dG =

Comparing these two equations shows that

(:t dT () T dp.- S,

(10.40)

(10.41) (10.42)

and Because of the importance of the Gibbs energy, Eqs. (10.41) and (10.42) contain two of the most important pieces of information in thermodynamics. Again, since the entropy of any substance is positive, the minus sign in Eq. (10.41) shows that increase in temperature decreases the Gibbs energy if the pressure is constant. The rate of decrease is greater for gases, which have large entropies, than for liquids or solids, which have small entropies. Because V is always positive, an increase in pressure increases the Gibbs energy at con stant temperature, as shown by Eq. (10.42). The larger the volume of the system the greater is the increase in Gibbs energy for a given increase in pressure. The comparatively large volume of a gas implies that the Gibbs energy of a gas increases much more rapidly with pressure than would that of a liquid or a solid. The Gibbs energy of any pure material is conveniently expressed by integrating Eq. (10.42) at constant temperature from the standard pressure, p O = 1 atm, to any other pressure p :

f dG = fP Vdp,orpO pO

p

G - GO =

fP V dp,pO

(10.43)where G O (T) is the Gibbs energy of the substance under 1 atm pressure, the s tandard Gibbs energy, which is a function of temperature. If the substance in question is either a liquid or a solid, the volume is nearly independ ent of the pressure and can be removed from under the integral sign ; then (liquids and solids).

(10.44)

21 4

Sponta n e ity a n d Eq u i l i br i u m

Since the volume of liquids and solids is small, unless the pressure is enormous, the second term on the right of Eq. (10.44) is negligibly small ; ordinarily for condensed phases we will write simply

(10.45)and ignore the pressure dependence of G. The volume of gases is very much larger than that of liquids or solids and depends greatly on pressure ; applying Eq. (10.43) to the ideal gas, it becomes

G

= GO(T) + fP nRT dp, ppO

. n

= GO(T) n

+

atm . RT In p 1 atm

It is customary to use a special symbol, fl, for the Gibbs energy per mole ; we define

( )

fl fl

=G n

- .

(10.46) (10.47)

Thus for the molar Gibbs energy of the ideal gas, we have As in Section 9.1 1, the symbol p in Eq. (10.47) represents a pure number, the number which when multiplied by 1 atm yields the value of the pressure in atmospheres. The logarithmic term in Eq. (10.47) is quite large in most circumstances and cannot be ignored. From this equation it is clear that at a specified temperature, the pressure describes the Gibbs energy of the ideal gas ; the higher the pressure the greater the Gibbs energy (Fig. 10.1). It is worth emphasizing that if we know the functional form of G(T, p), then we can obtain all other thermodynamic functions by differentiation, using Eqs. (10.41) and (10.42), and combining with definitions. (See Problem 10.29.)/.1 _ /.1 0

= flO(T) + RT In p.

2RT

RT

O r------+----r- P

- RT

- 2RT

F i g u re 1 0. 1 G i bbs energy of i d e a l gas as a fu n ction of press u re .

The G i b bs E n e rg y of R e a l G ases

21 5

1 0 . 9 T H E G I B B S E N E R G Y O F R EA L G A S E S

The functional form o f Eq. (10.47) i s particularly simple and convenient. It would be helpful if the molar Gibbs energy of real gases could be expressed in the same mathemat ical form. We therefore "invent" a function of the state that will express the molar Gibbs energy of a real gas by the equation

(10.48)The function f is called the fugacity of the gas. The fugacity measures the Gibbs energy of a real gas in the same way as the pressure measures the Gibbs energy of an ideal gas. An invented function such as the fugacity has little use unless it can be related to measurable properties of the gas. Dividing the fundamental equation (10.22) by n, the number of moles of gas, and restricting to constant temperature, dT = 0, we obtain for the real gas dJ1 = V dp, while for the ideal gas dJ1id = Vid dp, where V and Vid are the molar volumes of the real and ideal gases, respectively. Subtracting these two equations, we obtain d(J1 - J1id) = ( V - Vid) dp. Integrating between the limits p * and p yields

But by Eq. (10.47), J1id = J10( T) + R T In p, and by the definition of f, Eq. J10( T ) + R T hi.j. Using these values for J1 and J1id, Eq. (10.49) becomes R T(lnf - ln p) lnf =

fP*( V - Vid) dp. P We now let p * -+ The properties of any real gas approach their ideal values as the pres sure of the gas approaches zero. Therefore, as p * 0, J1 * -+ J1 * id . The equation becomes (10.49) J1 - J1id foP (V - j7id) dp.o. =-+

(J1 - J1id) - (J1 * - J1 * id) =

(10.48), J1 =

The integral in Eq. (10.50) can be evaluated graphically ; knowing V as a function of pressure, we plot the quantity (V - V id)/R T as a function of pressure. The area under the curve from p = 0 to p is the value of the second term on the right of Eq. (10.50). Or, if V can be expressed as a function of pressure by an equation of state, the integral can be evaluated analytically, since Vid = R T/p. The integral can be expressed neatly in terms of the compressibility factor Z; by definition, V = Z V id . Using this value for V, and V id = R T/p, in the integral of Eq. (10.50), it reduces toInf = In p +

f: d) dp ; 1 (P - - . d In p + (V - V I ) dp. R T Jo(V - V i

=

(10.50)

(P (Z p- 1) dp. Jo

(10.51)

The integral in Eq. (10.51) is evaluated graphically by plotting (Z - 1)/p against p and measuring the area under the curve. For gases below their Boyle temperatures, Z - 1 is negative at moderate pressures, and the fugacity, by Eq. (10.51), will be less than the pressure. For gases above their Boyle temperatures, the fugacity is greater than the pr ssure.

21 6

S ponta n e ity a n d Eq u i l i br i u m

The Gibbs energy of gases will usually be discussed as though the gases were ideal, and Eq. (10.47) will be used. The algebra will be exactly the same for real gases ; we need only replace the pressure in the final equations by the fugacity, keeping in mind that the fugacity depends on temperature as well as pressure.1 0 . 1 0 T E M P E RAT U R E D E P E N D E N C E O F T H E G I B B S E N E R G Y

The dependence of the Gibbs energy on temperature i s expressed in several different ways for convenience in different problems. Rewriting Eq. (10.41), we have

From the definition G =

H - TS, we obtain

(:t

= - So-S =

(10.52)

(G - H)/T, and Eq. (10.52) becomes G-H (10.53) T '

a form that is sometimes useful. Frequently it is important to know how the function By the ordinary rule of differentiation, we obtain

Using Eq.

(10.52), this becomes

(O(G/T) aT

p

=

OG T aT

which reduces to

(O(G/T) aT (O(G/T) aTp

( )

G/T depends on temperature.

p

_

G. ' T2

TS + G T2p

(10.54) (10.54) by (10.55)

the Gibbs-Helmholtz equation, which we use frequently. Since delfT) = - (1/T 2 ) dT, we can replace aT in the derivative in Eq. T 2 o(1/T) ; this reduces it to-

which is another frequently used relation. Any of Eqs. (10.52), (10.53), (10.54), (10.55) are simply different versions of the funda mental equation, Eq. (10.52). We will refer to them as the first, second, third, and fourth forms of the Gibbs-Helmholtz equation.Q U E STI O N S10.1

(O(G/T) o(1/T)

p

=H

,

For what sort of experimental conditions is (a) A or (b) G the appropriate indicator of spontaneity? 10.2 The second law states that the entropy of the universe (system and surroundings) increases in a spontaneous process : llSsyst + llSsurr ;?: O. Argue that, at constant T and p, llSsurr is related to the system enthalpy change by llSsurr = llHsysJT. Then argue that Eq. (10. 17) follows, where G is the system Gibbs energy.-

P ro b l ems

21 7

10.3 10.4

Discuss the meaning of the term " spontaneous " in thermodynamics.

Construct a I1H and I1S table, including the four possibilities associated with the two possible signs for each of I1H and I1S. Discuss the resulting sign of I1G and the spontaneity of the process. 10.5 The endothermic process of fo ming a solution of salt (NaCl) and water is spontaneous at room temperature. Explain how this is possible in terms of the higher entropy of the ions in solution compared to that of ions in the solid. Is the increase of Jl with increasing p for an ideal gas an enthalpy or an entropy effect ? 10.7 Explain why Eqs. (10. 1 7) and (10.47) do not imply that an ideal gas at constant temperature will spontaneously reduce its own pressure.10.6

PROBLEMS10.1 10.2

Using the van der Waals equation with the thermodynamic equation of state, evaluate (a Uja Vh for the van der Waals gas.

By integrating the total differential dU for a van der Waals gas, show that if Cv is a constant, U U' + Cv T - najr, where U' is a constant of integration. (The answer to Problem 10.1 is needed for this problem.) 10.3 Calculate I1U for the isothermal expansion of one mole of a van der Waals gas from 20 dm 3 jmol to 80 dm 3 jmol ; if a 0.141 m 6 Pa mol- 2 (nitrogen) and if a 3 . 1 9 m 6 Pa mol- 2 (heptane).10.4

=

a) Find the value of (aSja Vh for the van der Waals gas. b) Derive an expression for the change in entropy for the isothermal expansion of one mole of the van der Waals gas from Vi to V2 . c) Compare the result in (b) with the expression for the ideal gas. For the same increase in volume, will the entropy increase be greater for the van der Waals gas or for the ideal gas ? Evaluate the derivative, (a Uja Vh , for the Berthelot equation and the Dieterici equation. a) Write the thermodynamic equation of state for a substance that follows Joule's law. b) By integrating the differential equation obtained in (a), show that at constant volume the pressure is proportional to the absolute temperature for such a substance.

=

=

10.5 10.6

As a first approximation, the compressibility factor of the van der Waals gas is given by a 1 + RT T ' From this expression and the thermodynamic equation of state show that (aHjaph (2ajR T). 10.8 Using the expression in Problem 10.7 for the compressibility factor, show that for the van der Waals gas10.7

b-

= (b - ) :T

=

10.9

10.10

At 700 K calculate I1H and I1S for the compression of ammonia from 0.1013 MPa to 50.00 MPa, using the Beattie-Bridgeman equation and the constants in Table 3.5.

Using the results in Problems 10.7 and 10.8, calculate I1H and I1S for an isothermal increase in pressure of CO 2 from 0. 100 MPa to 10.0 MPa, assuming van der Waals behavior ; a 0.366 m 6 Pa mol- 2 , 42.9 X 10 - 6 m 3 jmol. a) At 300 K. b) At 400 K. c) Compare with the ideal gas values.

b=

= - [; + (RR;)2 J (:;)

=

218

S ponta n e ity a n d E q u i l i b r i u m

,/ 10 . 1 1

10.1 2

10.13

10.14

10.15

10.16 10.17

10.18

10.19

1 0.20

10.21

R T 2 OZ ' - oT p pwhere f.1rr is the Joule-Thomson coefficient, and Z = p VJR T is the compressibility factor of the gas. Compare to Eq. (7.50). Using the value of Z for the van der Waals gas given in Problem 10.7, calculate the value of f.1rr . Show that f.1rr changes sign at the inversion temperature, Tinv = 2a/Rb. a) Show that Eq. (10.31) can be written in the form OU O(P/T) O (P/T) = T2 = av T oT v o(l/T) v b) Show that Eq. (10.30) can be written in the form OH O (V/T) O (V/T) = _ T2 = op T oT p o(l/T) p At 25 C calculate the value of M for an isothermal expansion of one mole of the ideal gas from 10 litres to 40 litres. By integrating Eq. (10.39) derive an expression for the Helmholtz energy of a) the ideal gas ; b) the van der Waals gas. (Don't forget the " constant " of integration !) Calculate I1G for the isothermal (300 K) expansion of an ideal gas from 5000 kPa to 200 kPa. Using the form given in Problem 10.7 for the van der Waals equation, derive an expression for I1G if one mole of gas is compressed isothermally from 1 atm to a pressure p. Calculate I1G for the isothermal expansion of the van der Waals gas at 300 K from 5000 kPa to ioo kPa. Compare with the result in Problem 10. 16 for O 2 for which a = 0. 138 m 6 Pa mol - 2 and b = 3 1 .8 x 1O- 6 m 3 /moL At 300 K one mole of a substance is subjected to an isothermal increase in pressure from 100 kPa to 1000 kPa. Calculate I1G for ea.ch substance in (a) through (d) and compare the numerical values. a) Ideal gas. b) Liquid water ; V = 18 cm 3 /moL c) Copper ; V = 7.1 cm 3 /moL d) Sodium chloride ; V = 27 cm 3 /moL Using the van der Waals equation in the form given in Problem 10.7, derive the expression for the fugacity of the van der Waals gas. From the definition of the fugacity and the Gibbs-Helmholtz equation, show that the molar enthalpy, H, of a real gas is related to the molar enthalpy of the ideal gas, HO, by O ln J H = HO R T 2 oT p and that the molar entropy, S, is related to the standard molar entropy of the ideal gas So byCp f.1rr_

Show that for a real gas

=

( )

( )

( )

[ ] [

_

] [

[ ]

]

_

( )

S = So - R ln J + T10.22

Show also from the differential equation for dG that V = R T(o ln f/dp) r . Combining the results o f Problems 10.20 and 10.21 show that the enthalpy o f the van der Waals gas is 2a H = HO + b p. RT_ _

[

(O:;)J )

(

P r o b lems

21 9

10.23

From the purely mathematical properties of the exact differentialdU=

Cv dT +

1 0.24

10.25

1 0 .26

1 0.27

1 0.28

10.29

./1 0.30

show that if (8 Uj8 V)y is a function only of volume, then Cv is a function only of temperature. By taking the reciprocal of both sides of Eq. (10.23) we obtain (8Sj8p)v = - (8 Vj8 Y)s . Using this equation and the cyclic relation between V, T, and S, show that C8Sj8p)v = KCv/rt.T. Given dU = Cv dT + [(Trt. - pK)jKJ d V, show that dU = [Cv + (T Vrt. 2 /K) - p Vrt.J dT + V(pK - Trt.) dp. Hint: Expand d V in terms of d T and dp. Using the result in Problem 10.25 and the data for carbon tetrachloride at 20 C : rt. = 12.4 x 10 - 4 K - 1 ; K = 103 x 10 6 atm - 1 ; density = 1 . 5942 gjcm 3 and M = 153.8 gjmol, show that near 1 atm pressure, (8 Uj8p)y ;:::; - V Trt.. Calculate the change in molar energy per atm at 20 C. Using the approximate value of the compressibility factor given in Problem 10.7, show that for the van der Waals gas : a) Cp - Cv = R + 2 apjR T 2 b) (8 Uj8p)y = - ajR T. [Hint: Refer to Problem 10.25.J c) (8Uj8 T)p = Cv + apjR T2 . Knowing that dS = (CiT) dT - Vrt. dp, show that a) (8Sj8p)v = KCiTrt.. b) (8Sj8V)p = CpjTVrt. . c) - (ljV)(Wj8p)s = Kjy , where y == CpjCv . By using the fundamental differential equations and the definitions of the functions, determine the functional form of S, V, H, U for a) the ideal gas, given that !l = !l0(Y) + R T ln p. b) the van der Waals gas, given that !l = !l 0(T) + R T ln p + (b - ajR T)p . Show that if Z = 1 + B(T)p, then f = peZ - 1 ; and that this implies that at low to moderate pressures f ;:::; pZ, and that p 2 = fPidea] ' (This last relation states that the pressure is the geo metric mean of the ideal pressure and the fugacity.)

( )8U 8V

T

d V,

-

Syste m s of Va r i a b l e C o m po s i t i o n ; C h e m i ca l E q u i l i b r i u m

1 1 .1

T H E F U N DA M E N TA L E Q U AT I O N

In our study so far we have assumed implicitly that the system is composed of a pure substance or, if it was composed of a mixture, that the composition of the mixture was unaltered in the change of state. As a chemical reaction proceeds, the composition of the system and the thermodynamic properties change. Consequently, we must introduce the dependence on composition into the thermodynamic equations. We do this first only for the Gibbs energy G, since it is the most immediately useful. For a pure substance or for a mixture of fixed composition the fundamental equation for the Gibbs energy is (1 1 . 1) d G = S dT + V dp. If the mole numbers, nb n 2 , . . . , of the substances present vary, then G = G( T p, n l , n 2 , . . ), and the total differential is, . -

dG =

where the subscript n i on the partial derivative means that all the mole numbers are constant in the differentiation, and the subscript nj on the partial derivative means that all the mole numbers except the one in that derivative are constant in the differentiation. For example, C 8G/8n 2 h , p , n j means that T, p, and all the mole numbers except n 2 are constant in the differentiation. If the system does not suffer any change in composition, then

)p, n, (G) T, n, (:nGi ) T, p, nj n i (:nG2 ) T, p, nj dn2 ( PdT + dp + d

+

+ . . . , ( 1 1 .2)

222

Systems of Va r i a b l e Composition

and so on, and Eq. (1 1 .2) reduces to

dG =

Comparison of Eq. ( 1 1 .3) with Eq. (1 1 . 1), shows thatp, n ,

To simplify writing, we define

(8G ) 8T

(:G ) T-S

p, n ,

dT +

(G) P.

T, n,

dp.

( 1 1 .3)

-

and

(1 1 .4a, b)

8G l1-i = 8 ni

In view of Eqs. ( 1 1 .4) and (1 1.5), the total differential of G in Eq. ( 1 1 .2) becomes Equation ( 1 1 .6) relates the change in Gibbs energy to changes in the temperature, pressure, and the mole numbers ; it is usually written in the more compact form

( )

T ,.p, nj

( 1 1 .5)

dG = - S dT + V dp + 11- 1 dn 1 + 11- 2 dn 2 + . . . . dG = - S dT + Vdp + L l1-i dni ' iPi

(1 1 .6)

( 1 1.7)

where the sum includes all the constituents of the mixture.1 1 . 2 T H E P R O P E RT I E S O F

If a small amount of substance i, dni moles, is added to a system, keeping T, p, and all the other mole numbers constant, then the increase in Gibbs energy is given by Eq. (1 1 .7), which reduces to dG = l1-i dn i ' The increase in Gibbs energy per mole of (he substance added is

This equation expresses the immediate significance of l1- i ' and is simply the content of the definition of l1-i in Eq. (1 1 . 5). For any substance i in a mixture, the value of l1-i is the increase in Gibbs energy that attends the addition of an infinitesimal number of moles of that substance to the mixture per mole of the substance added. (The amount added is restricted to an infinitesimal quantity so that the composition of the mixture, and therefore the value of l1-i ' does not change.) An alternative approach involves an extremely large system, let us say a roomful of a water solution of sugar. If one mole of water is added to such a large system, the composition of the system remains the same for all practical purposes, and therefore the I1-H20 of the water is constant. The increase in Gibbs energy attending the addition of one mole of water to the roomful of solution is the value of I1-H20 in the solution. Since l1- i is the derivative of one extensive variable by another, it is an intensive property of the system and must have the same value everywhere within a system at equilibrium. Suppose that l1- i had different values, 11-1 and I1-f, in two regions of the system, A and B. Then keeping T, p, and all the other mole numbers constant, suppose that we transfer dni moles of i from region A to region B. For the increase in Gibbs energy in the two

(8Gi) 8n

T , p, n)

=

l1-i '

The G i b bs E n e rgy of a M ixtu re

223

regions, we have from Eq. (1 1 .7), dGA = ,4( - dn i), and dGB = flf dn i , since + dn i moles go into B and - dn i moles go into A. The total change in Gibbs energy of the system is the sum dG = dGA + dGB , or Now if flf is less than 111 , then dG is negative, and this transfer of matter decreases the Gibbs energy of the system ; the transfer therefore occurs spontaneously. Thus, substance i flows spontaneously from a region of high fl i to a region of low fli ; this flow continues until the value of fli is uniform throughout the system, that is, until the system is in equi librium. The fact that fli must have the same value throughout the system is an important equilibrium condition, which we will use again and again. The property fli is called the chemical potential of the substance i. Matter flows spontaneously from a region of high chemical potential to a region of low chemical potential just as electric current flows spontaneously from a region of high electrical potential to one of lower electrical potential, or as mass flows spontaneously from a position of high gravitational potential to one of low gravitational potential. Another name frequently given to fli is the escaping tendency of i. If the chemical potential of a component in a system is high, that component has a large escaping tendency, while if the chemical potential is low, the component has a small escaping tendency.1 1 . 3 T H E G I B B S E N E R G Y O F A M IXT U R E

dG = (flf - fl1 ) dni '

The fact that the l1i are intensive properties implies that they can depend only on other intensive properties such as temperature, pressure, and intensive composition variables such as the mole ratios, or the mole fractions. Since the fli depend on the mole numbers only through intensive composition variables, an important relation is easily derived. . Consider the following transformation :

Initial StateT, pSubstances Mole numbers Gibbs energy

Final StateT, p1

2 3 0 0 0 G=O1

2 3 n l n2 n3 G

We achieve this transformation by considering a large quantity of a mixture of uniform composition, in equilibrium at constant temperature and constant pressure. Imagine a small, closed mathematical surface such as a sphere that lies completely in the interior of this mixture and forms the boundary that encloses our thermodynamic system. We denote the Gibbs energy of this system by G* and the number of moles of the ith species in the system by nf , We now ask by how much the Gibbs energy of the system increases if we enlarge this mathematical surface so that it encloses a greater quantity of the mixture. We may imagine that the final boundary enlarges and deforms in such a way as to enclose any desired amount of mixture in a vessel of any shape. Let the Gibbs energy of the enlarged system be G and the mole numbers be n i ' We obtain this change in

224

Systems of Va r i a b l e C o m position

Gibbs energy by integrating Eq. (1 1.7) at constant T and p ; that is,

dG = L f.1i dni ; i n 't G - G* = I f.1 (ni - n t ). iG* ;

f

G

I ni

(1 1 .8)

The f.1i were taken out of the integrals because, as we have shown above, each f.1i must have the same value everywhere throughout a system at equilibrium. Now we allow our initial small boundary to shrink to the limit of enclosing zero volume ; then nt = 0, and G* = O. This reduces Eq. (1 1 .8) to

G = L n if.1i ' i

(11.9)

The addition rule in Eq. (1 1.9) is a very important property of chemical potentials. Knowing the chemical potential and the number of moles of each constituent of a mixture, we can compute, using Eq. (1 1.9), the total Gibbs energy, G, of the mixture at the specified temperature and pressure. If the system contains only one substance, then Eq. (11.9) reduces to G = nf.1, or

G f.1 = -. n

(1 1 . 10)

By Eq. (1 1.1 0), the f.1 of a pure substance is simply the molar Gibbs energy ; for this reason the symbol f.1 was introduced for molar Gibbs energy in Section 10.8. In mixtures, f.1i is the partial molar Gibbs energy of the substance i.1 1 .4 T H E C H E M I CA L P OT E N T i A L O F A P U R E I D EA L G A S

The chemical potential of a pure ideal gas is given explicitly by Eq.

(10.47) : (1 1 . 1 1)

f.1 = f.10( T) + R T ln p.

This equation shows that at a given temperature the pressure is a measure of the chemical potential of the gas. If inequalities in pressure exist in a container of a gas, then matter will flow from the high-pressure regions (high chemical potential) to those of lower pressure (lower chemical potential) until the pressure is equalized throughout the vessel. The equilibrium condition, equality of the chemical potential everywhere, requires that the pressure be uniform throughout the vessel. For nonideal gases it is the fugacity that must be uniform throughout the vessel ; however, since the fugacity is a function of temperature and pressure, at a given temperature equal values of fugacity imply equal values of pressure.1 1 . 5 C H E M I CA L P O T E N T I A L O F A N I D EA L G A S I N A M IXT U R E O F I D EA L G A S E S

Consider the system shown in Fig. 1 1.1. The right-hand compartment contains a mixture of hydrogen under a partial pressure PH2 and nitrogen under a partial pressure P N2 ' the total pressure being P = PH2 + PN2 ' The mixture is separated from the left-hand side by a palladium membrane. Since hydrogen can pass freely through the membrane, the left-hand side contains pure hydrogen. When equilibrium is attained, the pressure of the pure hydrogen on the left-hand side is equal by definition to the partial pressure of

C h e m i c a l Pote n t i a l of Gas i n a M ixtu re

225

Pure Hz

Palladium memb rane

i

F i g u re 1 1 . 1 C h e m i c a l potential o f a gas i n a m ixtu re .

the hydrogen in the mixture (see Section 2.8). The equilibrium condition requires that the chemical potential of the hydrogen must have the same value in both sides of the vessel :

,uH2(pur e) ,uH2 (mix) ' The chemical potential of pure hydrogen under a pressure PH2 is, by Eq. (1 1 . 1 1),

=

,uH2(pure) ,uH2 (T) + R T ln PH2 ' Therefore in the mixture it must be that ,uH2(mix) ,uH2 (T) + R T ln PH2 'This equation shows that the chemical potential of hydrogen in a mixture is a logarithmic function of the partial pressure of hydrogen in the mixture. By repeating the argument using a mixture of any number of ideal gases, and a membrane* permeable only to sub stance i, it may be shown that the chemical potential of substance i in the mixture is given by (1 1 . 12) ,u i ,ur C T) + R T ln p i ' where P i is the partial pressure of substance i in the mixture. The ,u r(T) has the same significance as for a pure gas ; it is the chemical potential of the pure gas under 1 atm pressure at the temperature T. By using the relation P i Xi P , where Xi is the mole fraction of substance i in the mixture and P is the total pressure, for P i in Eq. (1 1 . 1 2), and expanding the logarithm, we obtain (1 1.13) ,ui f-l r(T) + R T ln P + R T ln Xi '

= =

=

P,

By Eq. (1 1 . 1 1), the first two terms in Eq. (1 1 . 1 3) are the ,u for pure i under the pressure so Eq. (1 1 . 1 3) reduces to (1 1 . 14) f-l i ,ui (pure) ( T, p) + R T ln x i

= =

Since Xi is a fraction and its logarithm is negative, Eq. (1 1 . 14) shows that the chemical potential of any gas in a mixture is always less than the chemical potential of the pure gas under the same total pressure. If a pure gas under a pressure P is placed in contact with a mixture under the same total pressure, the pure gas will spontaneously flow into the mixture. This is the thermodynamic interpretation of the fact that gases, and for that matter liquids and solids as well, diffuse into one another. The form of Eq. (1 1 . 1 4) suggests a generalization. Suppose we define an ideal mixture, or ideal solution, in any state of aggregation (solid, liquid, or gaseous) as one in which*

=

.

The fact that such membranes are known for only a few gases does not impair the argument.

226

Systems of Va r i a b l e C o m position

the chemical potential of every species is given by the expression (1 1 . 14a) f1i = f1i(T, p) + RT ln Xi ' In Eq. (1 1 . 14a) we interpret f1i(T, p) as the chemical potential of the pure species i in the same state of aggregation as the mixture ; that is, in a liquid mixture, f1 i (T, p) is the chemical potential, or molar Gibbs energy, of pure liquid i at temperature T and pressure p, and Xi is the mole fraction of i in the liquid mixture. We will introduce particular empirical evidence to justify this generalization in Chapter 1 3 .

1 1 . 6 G I B B S E N E R G Y A N D E N T R O PY O F M I XI N G

Since the formation of a mixture from pure constituents always occurs spontaneously, this process must be attended by a decrease in Gibbs energy. Our object now is to calculate the Gibbs energy of mixing. The initial state is shown in Fig. 1 1.2(a). Each of the com partments contains a pure substance under a pressure p. The partitions separating the substances are pulled out and the final state, shown in Fig. 1 1 .2(b), is the mixture under the same pressure p. The temperature is the same initially and finally. For the pure sub stances, the Gibbs energies are The Gibbs energy of the initial state is simply the sumG init ial = G I

+

G2

+ G3

=

n l f1 + n 2 f1 + n 3 f1 3=

=

L: n i f1 i i

The Gibbs energy in the final state is given by the addition rule, Eq. ( 1 1 .9) :Gfinal =

n l f1 1 + n 2 f12 + n 3 f1 3

L n i f1i ' i

n l(f1 1 - f1n + n 2 (f1 2 - f1D + n 3 (f1 3 - f13 ) = L n lf1i - f1i). i Using the value of f1i - f1 i from Eq. ( 1 1 . 14a), we obtain Ll Gm ix = RT(n l In Xl + n 2 In X 2 + n 3 In X 3 ) = RT L n i In X i ' i which can be put in a slightly more convenient form by the substitution n i = x i n, whereLl Gmix =T, pn1

The Gibbs energy of mixing, Ll Gm ix = Gfinal - Ginit ial , on inserting the values of Gfinal and G initial , becomes

T, pn2

T, pn3

T, p

(a)F i g u re 1 1 . 2

(b)Free e nergy of m i x i n g . ( a ) I n itial state. ( b ) F i n a l state.

G i b bs E n e rgy a n d Entropy of M ix i n g

227

n is the total number of moles in the mixture, and Xi is the mole fraction of i. Then (1 1.15)which is the final expression for the Gibbs energy of mixing in terms of the mole fractions of the constituents of the mixture. Every term on the right-hand side is negative, and so the sum is always negative. From the derivation, it can be seen that in forming an ideal mixture of any number of species the Gibbs energy of mixing will be

fiGmix =

nRT I Xi In Xi 'i

(1 1. 16)

If there are only two substances in the mixture, then if X l = X , X 2 = 1 - X , Eq. (1 1.16) becomes fiGmix = nRT [x In X + (1 - x) In (1 - x)]. (1 1.17) A plot of the function in Eq. (1 1.17) is shown in Fig. 1 1.3. The curve is symmetrical about X = l The greatest decrease in Gibbs energy on mixing is associated with the formation of the mixture having equal numbers of moles of the two constituents. In a ternary system, the greatest decrease in free energy on mixing occurs if the mole fraction of each substance is equal to t , and so on. Differentiation of Gmix = Gfinal - G ini tial , with respect to temperature, yields Smix directly, through Eq. (l1Aa) :

Differentiating both sides of Eq.

(0 m ) e ) C ) (0 )ixp, n ,

=

al

p, n,

-

G ial

p, n ,

- (Sfinal - S initial) ;

GmiX aT

p , n,

(1 1. 18)

so that Eq.

(1 1 . 1 8) becomes

(0 T )fiG miX a

(1 1. 16) with respect to temperature, we havep , n,

=

nR L., Xi In Xi ' '\',

Smix =

- nR I Xi In Xi 'i

(1 1.19)

o

o

o

x

1

F i g u re 1 1 . 3

b i n a ry ideal m i xture.

I1Gm ;xlnR T for a

228

Systems of Va r i a b l e Composition

The functional form of the entropy of mixing is the same as for the Gibbs energy of mixing, except that T does not appear as a factor and a minus sign occurs in the expression for the entropy of mixing. The minus sign means that the entropy of mixing is always positive, while the Gibbs energy of mixing is always negative. The positive entropy of mixing corresponds to the increase in randomness that occurs in mixing the molecules of several kinds. The expression for the entropy of mixing in Eq. (11.19) should be compared to that in Eq. (9.75), which was obtained from the statistical argument. Note that N in Eq. (9.75) is the number of molecules, whereas in Eq. (1 1.19) n is the number of moles ; therefore different constants, R and k, appear in the two equations. A plot of the entropy of mixing of a binary mixture according to the equation

(1 1.20) i\Smix = - nR [x In x + (1 - x) In (1 - x)] is shown in Fig. 1 1 .4. The entropy of mixing has a maximum value when x = l Using x = ! in Eq. (11.20), we obtain for the entropy of mixing per mole of mixture i\Smixin = - R(! In ! + ! In !) = - R In ! = + 0.693R = 5.76 J/K mol.In a mixture containing only two substances, the entropy of mixing per mole of the final mixture varies between 0 and 5.76 J/K, depending on the composition. The heat of mixing can be calculated by the equation

(1 1.21) using the values of the Gibbs energy and entropy of mixing from Eqs. (1 1.16) and (11.19). This reduces Eq. (11.21) to nR T I Xi In Xi = i\Hmix + nR T I Xi In Xi 'i i i\Hmix = O. There is no heat effect ssociated with the formation of an ideal mixture. Using the result that i\Hmix = 0, Eq. (1 1.21) becomeswhich becomes

(11.22)

(1 1.23) - i\ Gmix = Ti\S miX " Equation (1 1.23) shows that the driving force, - i\ Gmix , that produces the mixing is entirely an entropy effect. The mixed state is a more random state, and therefore is a more

o/!,.Sm i.!nR for a

x

1

F i g u re 1 1 .4

b i n a ry ideal m i xtu re.

C h e m i c a l Eq u i l i b r i u m i n a M ixtu re

22 9

not large, in nonideal mixtures for which the heat of mixing is not zero, the heat of mixing must either be negative or only slightly positive if the substances are to mix spontaneously. If the heat of mixing is more positive than about 1300 to 1600 JImol of mixture, then .1G mix is positive, and the liquids are not miscible but remain in two distinct layers. The volume of mixing is obtained by differentiating the Gibbs energy of mixing with respect to pressure, the temperature and composition being constant,. .1 v.mIx =

probable state. If the value of 5.76 J/K mol is used for the entropy of mixing, then at T = 300 K, .1Gmix = - (300 K) (5.76 J/K mol) = - 1730 J/mol. Thus the Gibbs energy of mixing of an ideal binary mixture ranges from 0 to - 1730 JImo!. Since - 1730 JImol is

However, inspection of Eq. (1 1. 16) shows that the Gibbs energy of mixing is independent of pressure, so the derivative is zero ; hence, Ideal mixtures are formed without any volume change.1 1 . 7 C H E M I CA L E Q U I LI B R I U M I N A M I XT U R E

(13 i3p )T,.1Gmi

n,

.1 Vmix =

o.

(1 1.24)

Consider a closed system at a constant temperature and under a constant total pressure. The system consists of a mixture of several chemical species that can react according to the equation

(11.25)where the Ai represent the chemical formulas of the substances, while the V i represent the stoichiometric coefficients. This is the notation used in Sec. 1.7.1 for chemical reactions. It is understood that the Vi are negative for reactants and positive for products. We now inquire whether the Gibbs energy of the mixture will increase or decrease if the reaction advances in the direction indicated by the arrow. If the Gibbs energy decreases as the reaction advances, then the reaction goes spontaneously in the direction of the arrow ; the advance of the reaction and the decrease in Gibbs energy continue until the Gibbs energy ofthe system reaches a minimum value. When the Gibbs energy of the system is a minimum, the reaction is at equilibrium. If the Gibbs energy of the system increases as the reaction advances in the direction of the arrow, then the reaction will go spon taneously, with a decrease in Gibbs energy, in the opposite direction ; again the mixture will reach a minimum value of Gibbs energy at the equilibrium position. Since T and p are constant, as the reaction advances the change in Gibbs energy of the system is given by Eq. (1 1.7); which becomes

d G = L J1.i dni (1 1.26) i where the changes in the mole numbers, dn i ' are those resulting from the chemicalreaction. These changes are not independent because the substances react in the stoichio metric ratios. Let the reaction advance by moles, where is the advancement of the reaction ; then the number of moles of each of the substances present is

(1 1.27)

230

Systems of Va r i a b l e Compos i t i o n

where the n? are the numbers of moles of the substances present before the reaction advanced by moles. Since the n? are constant, by differentiating Eq. (1 1 .27) we obtain Using Eq. ( 1 1 .28) in Eq. (1 1.26), we obtain

dn i = d

Vi

(1 1 .28)

dG =which becomes

( Vi lli) d(1 1 .29)

The derivative, (oG/ah, p , is the rate of increase of the Gibbs energy of the mixture with the advancement of the reaction. If this derivative is negative, the Gibbs energy of the mixture decreases as the reaction progresses in the direction indicated by the arrow, which implies that the reaction is spontaneous. If this derivative is positive, progress of the reaction in the forward direction would lead to an increase in Gibbs energy ofthe system ; since this is not possible, the reverse reaction would go spontaneously. If (aG/oh . p is zero, the Gibbs energy has a minimum value and the reaction is at equi librium. The equilibrium condition for the chemical reaction is then

and

(OG)I

o T . p , eq

= 0,

(1 1 .30)

(1 1 . 3 1) =0 eq The derivative in Eq. (1 1 .29) has the form of an increase of Gibbs energy, f1G, since it is the sum of the Gibbs energies of the products of the reaction less the sum of the Gibbs energies of the reactants. Consequently we._will write I1G for (oG/oh , p and call 11G the reaction Gibbs energy. From the above derivation it is clear that for any chemical reaction (1 1 .32) The equilibrium condition for any chemical reaction is (1 1 .33) The subscript eq is placed on the quantities in Eqs. (1 1 . 3 1) and ( 1 1 .33) to emphasize the fact that at equilibrium the values of the Il'S are related in the special way indicated by these equations. Since each Ili is IllT, p, n, n , . . . , ) the equilibrium condition deter mines e as a function of T, p, and the specified values of the initial mole numbers.1 1 . 8 T H E G E N E R A L B E H AV I O R O F G AS A F U N CT I O N O F

(2:: Vi lli)

Figure 1 1. 5a shows the general behavior of G as a function of in a homogeneous system. The advancement, , has a limited range of variation between a least value, v and a greatest value, g . At 1 ' one or more of the products has been exhausted, while at g one or more of the reactants has been exhausted, At some intermediate value, e > G

The G e n e r a l B e h a v i o r of G as a F u nct i o n of ';G

231

ac a .;

=0

F i g u re 1 1 . 5

w

G i bbs energy as a f u n ction of t h e adva n cement.

passes through a minimum. The value e is the equilibrium value of the advancement. To the left of the minimum, 8 G/8 is negative, indicating spontaneity in the forward direction, while to the right of the minimum, 8 G/8 is positive, indicating spontaneity in the reverse direction. Note that even though in the case illustrated the products have an intrinsically higher Gibbs energy than the reactants, the reaction does form some products. This is a consequence of the contribution of the Gibbs energy of mixing. At any composition the Gibbs energy of the mixture has the form

G=

L: n i Pi ' i

If we add and subtract pf(T, p), the chemical potential of the pure species i in each term of the sum, we obtain

G=

I n;(pf + P i - pf) = I ni pf (T, p) + I ni(P i - pD i i i( 1 1.34)

The first sum is the total Gibbs energy of the pure gases separately, Gpure ; the last sum is the Gibbs energy of mixing, Gmix ' The Gibbs energy of the system is given by The plot of G pure , G miX ' and G as a function of the advancement is shown in Fig. 1 1 .5b. Since Gpure depends on only through the n i ' each of which is a linear function of , we see that Gpure is a linear function of . The minimum in G occurs at the point where Gmix decreases as rapidly as G pure increases ; by differentiating,

At equilibrium

( ) ( ) +( ) ( ) ( )8G 8T, p

=

8 Gpure 8

T, p

8 'Gmix 8

T, p

.

8 Gpu re = 8 eq

_

8 Gmix . 8 eq

232

Systems of Va r i a b l e C o m position

This condition can be established geometrically by reflecting the line for Gpure in the horizontal line 00, to yield the line the point of tangency of the line parallel to with the curve for AGmix yields the value of the advancement at equilibrium. Equation (11.34) is correct for any equilibrium in a homogeneous ' system. Equation (1 1.34) is, in fact, formally correct for any equilibrium, but unless at least one phase is a mixture, the term Gmix , will be zero and only the first term, Gpure , will appear. Equation (11.34) shows that a system approaches the equilibrium state of minimum Gibbs energy by forming substances of intrinsically lower Gibbs energy ; this makes Gpure small. It also lowers its Gibbs energy by mixing the reactants and products. A compromise is reached between a pure material having a low intrinsic Gibbs energy and the highly mixed state.

OA,

OA;

O'A',

1 1 . 9 C H E M I CA L E Q U I LI B R I U M I N A M IXT U R E O F I D EA L G A S E S

I t has been shown, Eq. (11.12), that the Jl o f an ideal gas i n a gas mixture i s given by where Pi is the partial pressure of the gas in the mixture. We use this value of Jli in Eq. (11.29) to compute the AG for the reaction. aA +Jli

= Jlf + R T ln Pi '

(1 1.35) c5D

f3B

yC +

where A, B, C, and D represent the chemical formulas of the substances, while a, 13, y, and c5 represent the stoichiometric coefficients. Then

AGLet

= YJle + yRT ln Pc + c5Jl'tJ + c5R T In D aJlA - aR T In f3Jl - f3R T In PB = YJle + c5Jl'tJ - (aJlA + f3Jl) + R T[yPIn Pc + c5 ln PD (aPAIn PA + 13 In PB)] . '-

(11.36)AGO is the standard reaction Gibbs energy. Then, combining the logarithmic terms, AG

The argument of the logarithm is called the proper quotient of pressures ; the numerator is the product of partial pressures of the chemical products each raised to the power of its stoichiometric coefficient, and the denominator is the product of the partial pressures of the reactants, each raised to the power of its stoichiometric coefficient. Ordinarily the quotient is abbreviated by the symbol Qp :

= AGO + R T ln PCPD . PAP'

1

{)

(1 1.37)

(11.38)This reduces Eq. (11.37) to

(1 1.39)The sign of AG is determined by the sign and magnitude of In Q p , since at a given temperature AGo is a constant characteristic of the reaction. If, for example, we compose the mixture so that the partial pressures of the reactants are very large, while those of the products are very small, then Qp will have a small fractional value, and In Qp will be a

C h e m i c a l Eq u i l i br i u m i n a M ixtu re of Ideal G ases

233

large negative number. This in turn will make tJ.G more negative and iJilcrease the tendency for products to form. At equilibrium, tJ.G = 0, and Eq. (1 1 .37) becomes 0=tJ.Go

+ R T ln

(Pd(PD)

(PA)(PB)e

;

,

(1 1 .40)

where the subscript e indicates that these are equilibrium partial pressures. The quotient of equilibrium partial pressures, is the pressure equilibrium constant Kp :Kp

= (Pd(Po)

(PA)(PB) .

(1 1 .41)

Using the more general notation, we put the value of Ili from Eq. ( 1 1 . 35) in Eq. (1 1 .29) to obtaintJ.G

=

which can be written, But

tJ.G

()i

T, p

=

v ;(lli + R T ln P i), ,itJ.Go,

= I V i lli + R T I V i In P i '

I V i lli =i

(1 1 .36a)

the change in the standard reaction Gibbs energy, and V i In P i = In pii ; thus the equation becomes . tJ.G = tJ.Go + R T I In pii (1 1 .37a)i

This continued product,

But a sum of logarithms is the logarithm of a product : In p '? + In p2 + In p;3 + . . , = In (p lp2p;3 . . J

IT P iV i - pVIl pV22pV33 " ' ,i Qp =

is called the proper quotient of pressures, Qp .

IT p i ii

(1 1.38a)

Note that since the V i for the reactants are negative, we have for the reaction in questionVz

= - [3,_

and

PB PO y b Q p - PA aPB pPePD - -a-p Po Pe_

y

b

( l 1 .38b) (1 1 .41a) ( 1 1.42)

Correspondingly, K p can be written asKp =

IT (P i) ii

Equation ( 1 1 .40) becomes

234

Systems of Va r i a b l e C o m positi o n

, The quantity f1Go is a combination of flo S, each of which is a function only of temperature ; therefore f1Go is a function only of temperature, and so K p is a function only of temperature. From a measurement of the equilibrium constant of the reaction f1GO can be calculated using Eq. ( 1 1 .42). This is the way in which the value of f1Go for any reaction is obtained.II!! EXAMPLE 1 1 . 1

For the reaction !N ig)

+ !Hig) NHig), the equilibrium constant is 6.59 x 10 - 3 at 450 C. Compute the standard reactionGibbs energy at 450 C.

Solution.

- (8.314 J/K mol) (723 K) In (6.59 x 10 3 ) = - (6010 J/mol) ( - 5.02) = + 30 200 J/mo!. Since this is the formation reaction for ammonia, it follows that 30 200 J/mol is the standard Gibbs energy of formation of ammonia at 450 C.

f1GO =

1 l . 1 0 C H E M I CA L E Q U I LI B R I U M I N A M I XT U R E O F R EA L G A S E S

I f the corresponding algebra were carried out for real gases using Eq. (10.48), the equation equivalent to Eq. (1 1 .41) is ( 1 1 .43)

and, corresponding to Eq. (1 1.42), For real gases, it is K J rather than K p that is a function of temperature only. 1 1 . 1 1 T H E E Q U I LI B R I U M C O N STA N T S ,Kx

f1GO =

- R T ln K J .Kc

(1 1.44)

AN D

It is sometimes advantageous to express the equilibrium constant for gaseous systems in terms of either mole fractions, Xi ' or concentrations, ci , rather than partial pressures. The partial pressure, P i ' the mole fraction, and the total pressure, p, are related by P i = Xi P , Using this relation for each of the partial pressures in the equilibrium constant, we obtain from Eq. (1 1 .41)Kp = Cp (Pd(PD) = (X mXD P) = (Xd(XD) P y H - a p . (PA)(PB) (XAP)(XB P) (XA)(XB) (Xd(XD) (XA)(XB) '

The mole fraction equilibrium constant is defined byKx =

(1 1 .45) (1 1 .46)

Then where f1v = L Vi is the sum of stoichiometric coefficients on the right-hand side of the chemical equation minus the sum of the coefficients on the left-hand side. Rearranging Eq. (1 1 .46), we obtain Kx = Kp p - b. v . Since Kp is independent of pressure, Kx will depend on pressure unless f1v is zero.

Sta n d a rd G i bbs E n e rg i es of Formation

235

Keep in mind that in Kp the P i are pure numbers-abbreviations for the ratio Pi/(l atm) -which we will write as P i/po ; see the discussion of Eqs. (9.52), (9.53), and (10.47). It follows that the pressure in Eq. (1 1 .46) is also a pure number ; it is an abbreviation for p/po p/(l atm). In a similar way, since the partial pressure of a gas is given by P i n i R T/V and the concentration is C i ni/V, we obtain P i Ci R T. Introducing the standard pressure explicitly, we have

=

=

=

=

P Before we put this in K p it is useful to have Ci in a dimensionless ratio, so we multiply and divide by a standard concentration, co. Then we have(1 1 .47) Since we have a ratio of concentrations, it follows that

.

where the Ci and CO are concentrations in moljL, whereas the Ci and CO are the corresponding concentrations in mol/m 3 , the SI unit of concentration. As before, we will abbreviate pdpo as P i and c dc o cd(1 moljL) as ci ; then we have

=

Pi

in which P i and Ci are to be understood as the pure numbers equal to the ratios p;/(l atm) and c;/(l moljL). If we insert these values of P i in Kp by the same argument that we used to obtain Eq. ( 1 1 .46), we find COR T (1 1 .49) K p - Kc -pO where Kc is a quotient of equilibrium concentrations ; Kc is a function of temperature only. Since the standard concentration was CO 1 moljL, the corresponding value of C O 10 3 mol/m 3 ; thus cOR T (10 3 moljm 3 ) (8.31441 J/K mol)T 0.0820568 T/K' 101 325 Pa pO and we have RT ( 1 1 . 50) Kp Kc Kc(O.0820568 T/KY'. v 101.325 J/mo}

( ) dV _=

= c{ O;T)

(1 1 .48)

=

=

Note that the quantity in the parentheses is dimensionless, as are K p and Kc .1 1 . 1 2 STA N D A R D G I B B S E N E R G I E S O F F O R M AT I O N

= (

) dV =

=

Having obtained values of I1Go from measurements of equilibrium constants, it is possible to calculate conventional values of the standard molar Gibbs energy 11 of individual compounds. Just as in the case of the standard enthalpies of substances, we are at liberty

236

Systems of Va r i a b l e Compos i t i o n

to assign a value of zero to the Gibbs energy of the elements in their stable state of aggrega tion at 25 C and 1 atm pressure. For example, at 25 C For the formation reaction of a compound such as CO, we have C(graphite) .u(Br z , 1) =

0,

.u(S, rhombic) = CO(g),

O.

+ 1 0 z (g)

-------+

.u(CO, g) - [a O(C, graphite) + !.u(O z , g)]. Since .u(C, graphite) = 0 and .u(O z , g) = 0 by convention, we have=

!1G'}

Consequently, the standard Gibbs energy of formation of any compound is equal to the conventional standard molar Gibbs energy of that compound. Some values of the standard Gibbs energy of formation at 25 C are given in Table A- V . It is always possible to relate the composition of an equilibrium mixture to the equilibrium value of the advancement, e ' the initial mole numbers, nf, and the stoichio metric coefficients, Vi ' Two examples will be discussed.III EXAMPLE 1 1 .2

!1G'} = .u(CO, g)

(1 1 .51)

The dissociation of dinitrogen tetroxide. N z 0 4 (g):;::==::::':

2 NO z Cg)

This equilibrium can be easily studied in the laboratory through a measurement of the vapor density of the equilibrium mixture. In the following formulation the various quantities are listed in columns under the formulas of the compounds in the balanced chemical equation. Let n be the initial number of moles of N z 0 4 , e the equilibrium advancement, and lXe the fraction dissociated at equilibrium lXe = e/n. N z O ig) -1 n n - e n - e n + e 1 - lXe + lXe - IX 1 + IX: P--

2 NO z (g) +2

Stoichiometric coefficient Initial mole numbers, nf Equilibrium mole numbers, ni Total number of moles, n = n + e Mole fractions, Xi or, since lXe = e/n, the Xi are Partial pressures, Pi =XiP

0 o + 2e2e n + e 2IXe 1 + lXe 2IXe 1 + lXe

U sing these values of the partial pressures, we obtain2 Kp = P0 = PN204

C )

I

(

--

)P(1 1.52)

6-!: rP

1 - lXe 1 + lXe

J7

Sta n d a rd G i bbs E n e rg i es of Format i o n

231

It is clear that as p --+ 0, exe --+ 1, while as p --+ 00 , ex e --+ O. This is what would be expected from the LeChatelier principle. At moderately high pressures, Kp 4p and exe = 1K; / 2/p 1 / 2 , approximately.Iiil EXAMPLE 1 1 .3

By the ideal gas law, p V = nRT, where n = (1 + ex e )n. Thus p V_ (L+ exe)R T. But _I1 n = w/M, where w is the mass of gas in the volume V and Mis- the molarmass of N 2 0 4 . Thus, if we know p, T, V, and w we can calculate ex e and then, using Eq. (11.52), we can obtain Kp . A measurement of exe at any pressure p suffices to determine K p . From Kp ' I1Go can be calculated. The dependence of ex e on the pressure can be obtained explicitly by solving Eq. (1 1.52) for ex e : Kp exe = Kp + 4p

J'

The ammonia synthesis. Suppose we mix one mole of N 2 with 3 moles of H 2 (the stoichiometric ratio) and consider the equilibrium : N 2 (g) Stoichiometric coefficients Initial mole numbers,

+ 3 H z (g)-3 3 3 - 3 3( 1 2(2 3 (1 2(2 -

2 NHig)

-1 1 1- 2 1- 2(2 - ) 1- 2(2 - ) P

n?

Total number of moles, n = 4 Mole fractions, X i Partial pressures, P i = xiP

Equilibrium mole numbers, n i

2 0 2 2 2(2 - ) 2p 2(2 - )

) ) ) ) P

We note immediately that PH2 = 3PN2 ; using these values in Kp , we get 2 3 KP = PNH Taking the square root, we havePN2 P2

or, using the partial pressures from the table,

Analysis of the mixture yields the value of XNH3 from which we can obtain the value of at equilibrium. From the experimental value of we can calculate Kp , and from that, I1Go. We can also formulate the expression in terms of PNH 3 and the total pressure. Since

238

Systems of Va r i a b l e C o m pos i t i o n

P

Then

= PN2 + PH2 + PNH3 and PH2 = 3 PN2 ' then P = 4 PN2 + PNH3 or PN2 = i(P - PNHJ16

From this relation, the partial pressure of NH 3 can be calculated at any total pressure. If the conversion to NH 3 is low, then P - PNH 3 p, and PNH 3 0.325K / 2p 2 , so that the partial pressure of ammonia is approximately proportional to the square of the pressure. If the reactants are not mixed originally in the stoichiometric ratio, the expression is more complex. A measurement of the equilibrium partial pressure of NH 3 at a given temperature and pressure yields a value of /}.Go for this reaction, which is twice the conventional standard molar Gibbs energy of NH 3 at this temperature. Note that we have suppressed the subscripts on e and (PNH , )e to avoid a cumbersome notation. We will usually omit the subscript except when it is needed to avoid confusion. It is to be understood that all the quantities in the equilibrium constant are equilibrium values.

=

1 1 . 1 3 T H E T E M P E R AT U R E D E P E N D E N C E O F T H E E Q U i li B R I U M C O N STA N T

The equilibrium constant can b e written as (1 1.53) Differentiating, we obtain d In Kp

dT

(1 1.54)

Dividing Eq. (1 1.36a) b y T, w e obtain

Differentiating, we have

d(/}'G O/T) dT

= I v. d(pf/T) dTi !

(1 1 .55)

where the pf are standard molar Gibbs energies of pure substances. Using molar values in the Gibbs-Helmholtz equation, Eq. (10.54), we have d(pf/T)/dT - Hf/T 2 . This relation reduces Eq. (1 1 .55) to

=

d(/}'G O/T) dT

=

_

_

1 . H? T2 I v! ! i

=

_

/}.HO T2 '

(1 1 .56)

since the summation is the standard enthalpy increase for the reaction, /}.Ho. Equation (1 1 . 56) reduces Eq. (1 1 . 54) to or 2.303 R T 2 ' (1 1 . 57)

Equation ( 1 1 .57) is also called the Gibbs-Helmholtz equation.

The Temperat u re Dependence of t h e Eq u i l i b r i u m Con sta nt

239

If the reaction is exothermic, !1Ho is negative, and the equilibrium constant decreases with increase in temperature. If the reaction is endothermic, !1Ho is positive ; then Kp increases with increase in temperature. Since an increase in the equilibrium constant implies an increase in the yield of products, Eq. ( 1 1 . 57) is the mathematical expression of one aspect of the LeChatelier principle. Equation ( 1 1 .57) can be expressed readily in a form convenient for plotting :

Since !1Ho is almost constant, at least over moderate ranges of temperature, the plot is often linear. If Kp is measured at several temperatures and the data plotted as In Kp versus liT, the slope of the line yields a value of !1Ho for the reaction through Eq. (1 1 .58). Conse quently, it is possible to determine heats of reaction by measuring equilibrium constants over a range of temperature. The values of the heats of reaction obtained by this method are usually not so precise as those obtained by precision calorimetric methods. However, the equilibrium method can be used for reactions that are not suited to direct calorimetric measurement. Later we will find that certain equilibrium constants can be calculated from calorimetrically measured quantities only. Having obtained values of !1Go at several temperatures and a value of !1Ho from the plot of Eq. (1 1.58), we can calculate the values of !1So at each temperature from the equation (1 1.59) The equilibrium constant can be written as an explicit function of temperature by integrating Eq. (1 1.57). Suppose that at some temperature To , the value of the equilibrium constant is (Kp)o and at any other temperature T the value is Kp : !1HO !1HO d(ln Kp) = - dT, In Kp In (Kp)o = ""j.{2 dT, 2

dT = !1Ho R d T ' R T2 !1Ho d logl o Kp d In Kp (1 1 .58) d( 1 /T) d(lIT) 2.303 R Equation (1 1 .58) shows that a plot of In Kp versus liT has a slope equal to - !1HolR . d In Kp =!1Ho_

()

fIri Kp

In.(Kp)o

T ITo R T

In Kp = In (Kp)o

If !1Ho is a constant, then by integrating, we have !1Ho 1 1 (1 1.61) In Kp = In (Kp)o - . R T To From the knowledge of !1H0 and a value of (Kp)o at any temperature To , we can calculate Kp at any other temperature. If, in Eq. (1 1 . 53), we set !1Go = !1Ho T !1So, we obtain !1Ho !1So (1 1.61a) In Kp = - + -

T !1HO dT. + ITo RT2

-

T ITo

T

(1 1 .60)

(

)

RT

R

This relation is always true. But if !1Ho is constant, then !1So must also be constant, and this equation is equivalent to Eq. ( 1 1 .61). (Note that constancy of !1Ho implies that !1C = 0; but if !1C = 0, then !1So must also be constant.)

240

Systems of Va r i a b l e Composition

If I:lHo is not a constant, it can ordinarily be expressed (see Section 7.24) as a power series in T :

+ A' T + B' T 2 + C'T 3 + . . . . Using this value for I:lHo in Eq. (1 1.60) and integrating, we obtain I:lH'O l 1 B T In Kp = In (Kp) o - - - + - In + R ( T - To ) R T To T RI:lHo = I:lH'O

-- (

)

(-) o/

(1 1.62)which has the general functional form In Kp = A T

+ B + C In T + DT + E T 2 + . . . ,

(1 1.63)

tabulated) from a measurement at some other (usually higher) temperature. To evaluate the constants, the values of I:lHo and the heat capacities of all the reactants and products must be known.1 1 . 1 4 e Q U I LI B R I A B ETW E E N I D EA L G A S E S A N D P U R E C O N D E N S E D P H AS E S

and E are constants. Equations having the general form of Eq. in which A, B, C, (1 1.63) are often used to calculate an equilibrium constant at 25 C (so that it can be

D,

If the substances participating in the chemical equilibrium are in more than one phase, the equilibrium is heterogeneous. If the substances are all present in a single phase, the equilibrium is homogeneous. We have dealt so far only with homogeneous equilibria in gases. If, in addition to gases, a chemical reaction involves one or more pure liquids or solids, the expression for the equilibrium constant is slightly different.1 1 .1 4.1 T h e l i mest o n e D eco m po s i t i o n

Consider the reaction CaO(s) The equilibrium condition is [,u(CaO, s)

+ CO 2 (g).q

the pure solids (and for pure liquids if they appear), because of the insensitivity of the Gibbs energy of a condensed phase to change in pressure, we have ,u(CaO, s) = ,u (CaO, s). The equilibrium condition becomeso

+ ,u(C0 2 , g) - ,u(CaC0 3 , s)] e = O . For each gas present, e.g., CO 2 , [,u(C0 2 , g)] e = ,u (CO Z ' g) + R T In (PC02 )e ' While forq

0 = ,u (CaO, s) = I:lGo

+ ,u (C0 2 , g) - ,u (CaC0 3 , s) + R T ln (PC02)e , (1 1.64)

+ R T In (Peo, )e '

E q u i l i b r i a B etween I d e a l G ases a n d P u re Condens d P hases

241

In this case, the equilibrium constant is simplyKp

= (PCO,)e .CaO(s) - 604.0 - 63 5.09 - 394.36 - 393.51

The equilibrium constant contains only the pressure of the gas ; however, the !J.Go contains the standard Gibbs energies of all the reactants and products. From the data in Table A-V, we find (at 25 C)

Substancep O/(kJ/mol) LlHjl(kllmol) Then for the reaction and

CaCOis) - 1 128.8 - 1 206.9

= - 604.0 - 394.4 - ( - 1 128.8) = BOA kllmol, LlHO = - 635 . 1 - 393.5 - ( - 1 206.9) = 178.3 kllmo!.!J.GOIn (pCO2 ) e 30 = (8.314 1J/K400 J/mol 1 5 K) mol) (298. (at 298 K). (Peo2 ) e = 1 .43 10- 2 3 atm_ _ _

The equilibrium pressure is calculated from Eq. ( 1 1.64). 52.60 ,.

x

Suppose we want the value at another temperature, 1 100 K. We use Eq. (1 1 .61) : 1 178 = - 52.60 - 8.314300 JImol (1 l00 K - 298.11 5 K) - 0. 1 7 , J/K mol (PCO,) 1 1 00 = 0.84 atm._ .

1 1 . 1 4 . 2 The D e c o m p o si t i o n o f M e rcu r i c O x i d e

Consider the reaction HgO(s)

Hg(l) 2 . Also The equilibrium constant is Kp (P0 2 ) / ll Go pO(Hg, 1) + ! p O(0 2 ' g) - p O(HgO, s) Then

:;:::=:==::

+ ! 0 2 (g)

=

=

= - pO(HgO, s) = 58.56 kllmol._ .

In (P02) e - -

_

(PO, ) e

=

58 560 J/mol - - 23.62 , (8.314 11K mol) (298. 1 5 K) 5.50 x 10 - 1 1 atm.

242

Systems of Va r i a b l e Composition

1 1 . 1 4 . 3 Va p o r i zat i o n E q u i l i b r i a

An important example o f equilibrium between ideal gases and pure condensed phases is the equilibrium between a pure liquid and its vapor : A(l) A(g). Let p be the equilibrium vapor pressure. Then

Using the Gibbs-Helmholtz equation, Eq. ( 1 1 .57), we have

Kp = p

and

I1Go

= flO(g) - !l0(l).

d In p dT

I1Hap RT 2 '

(1 1 .65)

which is the Clausius-Clapeyron equation ; it relates the temperature dependence of the vapor pressure of a liquid to the heat of vaporization. A similar expression holds for the sublimation of a solid. Consider the reaction A(s)

A(g) ;

and

where p is the equilibrium vapor pressure of the solid. By the same argument as above

d In p dT

(1 1 .66)

where I1Hub is the heat of sublimation of the solid. In either case, a plot of In p versus 1/T has a slope equal to - I1Ho/R and is nearly linear.* 1 1 . 1 5 T H E L E C H AT E LI E R P R I N C I P L E

It i s fairly easy t o show how a change in temperature o r pressure affects the equilibrium value of the advancement e of a reaction. We need only to determine the sign of the derivatives (8 e/8T)p and (8 e /8p)y . We begin by writing the identity

Since (8G/8h, p is itself a function of T, expression,

() T, p+=

=

I1G.

(1 1.67)

d

Using Eq. ( 1 1 .67) and setting

118

From the fundamental equation, (8 I1G/8T) = - 118 and (8 I1G/8p) = I1V, in which is the entropy change and 11 V is the volume change for the reaction. Thus

() 8 () dT :p () dp 88 () d. (8 2 G/8 2 ) Gil, 8 I1G (8G 8 I1G dT ---ap dp G d. d a[)=

p, and we may write the total differential+( 1 1.68)

Eq. (1 1 .68) becomes

= fiT

+

+

II

d

If we insist that these variations in temperature, pressure, and advancement occur while

()

= - LS dT +

I1V dp + Gil d.

The leChate l i e r P r i n c i p l e

243

keeping the reaction at equilibrium, then aG/a = = I1H/T, so the equation becomes equilibrium,o

115

= -

At equilibrium G is a minimum ; therefore G must be positive. At constant pressure, dp = 0, and Eq. (1 1.69) becomes At constant temperature, dT =

(11:)

0 and hence d(aG/a) = O. At (1 1.69)

(dT) eq +

11

V(dP)e q + G(d e).

0, and Eq. (11.69) becomes

( )T

a e aT p

I1H

( 1 1 70).

Equations (1 1.70) and (1 1.71) are quantitative statements of the principle of LeChatelier : They describe the dependence of the advancement of the reaction at equi librium on temperature and on pressure. Since G is positive, the sign of (a e/aT)p depends on the sign of I1H . If I1H is + , an endothermic reaction, then (ae/a T)p is + , and an increase in temperature increases the advancement at equilibrium. For an exothermic reaction, I1H is - , so (a e/a T)p is - ; increase in temperature will decrease the equilibrium advancement of the reaction. Similarly, the sign of (ae/ap)y depends on .1. V. If V is - , the product volume is less than the reactant volume and (ae/ap) y is positive ; increase in pressure increases the equilibrium advancement. Conversely, if Li Vis + , then (ae/ap) y is - ; increase in pressure decreases the equilibrium advancement. The net effect of these relations is that an increase in pressure shifts the equili brium to the low-volume side of the reaction while a decrease in pressure shifts the equilibrium to the high-volume side. Similarly an increment in temperature shifts the equilibrium to the high-enthalpy side, while a decrease in temperature shifts it to the low-enthalpy side. We may state the principle of LeChatelier in the following way. If the external con straints under which an equilibrium is established are changed, the equilibrium will shift in such a way as to moderate the effect of the change. For example, if the volume of a nonreactive system is decreased by a specified amount, the pressure rises correspondingly. In a reactive system, the equilibrium shifts to the low-volume side (if Li V =1= 0), so the pressure increment is less than in the nonreactive case. The response of the system is moderated by the shift in equilibrium position. This implies that the compressibility of a reactive system is much greater than that of a non reactive one (see Problem 1 1.39). Similarly, if we extract a fixed quantity of heat from a nonreactive system, the tempera ture decreases by a definite amount. In a reactive system, withdrawing the same amount of heat will not produce as large a decrease in temperature because the equilibrium shifts to the low-enthalpy side (if LiH =1= 0). This implies that the heat capacity of a reactive system is much larger than that of a nonreactive one (see Problem 1 1.40). This is useful if the system can be used as a heat-transfer or heat-storage medium. It must be noted here that there are certain types of systems that do not obey the LeChatelier principle in all circumstances (for example, open systems). A very general

( )a e ap

I1 V il Ge

(1 1.71)

11

244

Systems of Va r i a b l e Composition

validity has been claimed for the LeChatelier principle. However, if the principle does have such broad application, the statement of the principle must be very much more complex than that given here or in other elementary discussions. * 1 1 . 1 6 E Q U I LI B R I U M C O N STA N T S F R O M CALO R I M ET R I CM EAS U R E M E NTS . T H E T H I R D LAW I N ITS H I ST O R I CA L C O N T EXT

Using the Gibbs-Helmholtz equation, we can calculate the equilibrium constant of a reaction at any temperature T from a knowledge of the equilibrium constant at one temperature To and the I1Ho of the reaction. For convenience we rewrite Eq. (1 1.60) : T I1HO dT. In Kp = In (Kp) o + To RT 2 The I1Ho for any reaction and its temperature dependence can be determined by purely thermal (that is, calorimetric) measurements. Thus, according to Eq. (1 1 .60), a measure ment of the equilibrium constant at only one temperature together with the thermal measurements of I1Ho and I1Cp suffice to determine the value of Kp at any other tem perature. The question naturally arises whether or not it is possible to calculate the equilibrium constant exclusively from quantities that have been determined calorimetrically. In view of the relation I1Go = RT ln Kp , the equilibrium constant can be calculated if I1Go is known. At any temperature T, by definition,

I

-

(1 1.72)Since I1Ho can be obtained from thermal measurements, the problem resolves into the question of whether or not I1So can be obtained solely from thermal measurements. For any single substance

(11.73) where Sr is the entropy of the substance at temperature T; So , the entropy at 0 K, and SO-+ T is the entropy increase if the substance is taken from 0 K to the temperature T. The SO -+ T can be measured calorimetrically. For a chemical reaction, using Eq. (11.73)for each substance Putting this result into Eq. (1 1.72), we obtain Therefore

I1So = I1So + I1S 0 -+ T '-

I1Go = I1HoIn K =

T I1So+

-

T I1S0 -+ T '

I1So

R

I1So -+ T

R

_ 11HRT

(1 1.74)

Since the last two terms in Eq. (1 1.74) can be calculated from heat capacities and heats of reaction, the only unknown quantity is I1So , the change in entropy of the reaction at 0 K. In 1906, Nernst suggested that for all chemical reactions involving pure crystalline solids, I1So is zero at the absolute zero ; the N ernst heat theorem. In 1913, Planck suggested that the reason that I1So is zero is that the entropy of each individual substance taking part in such a reaction is zero. It is clear that Planck ' s statement includes the Nernst theorem.

C h e m i c a l R eactions a n c! the Entropy of t h e U n iverse

245

However, either one is sufficient for the solution of the problem of determining the equilibrium constant from thermal measurements. Setting LlSg = 0 in Eq. (1 1.74), we obtain (1 1 .75) where LlSo is the difference, at temperature T, in the third-law entropies of the substances involved in the reaction. Thus it is possible to calculate equilibrium constants from calorimetric data exclusively, provided that every substance in the reaction follows the third law. Nernst based the heat theorem on evidence from several chemical reactions. The data showed that, at least for those reactions, LlGo approached LlHo as the temperature decreased ; from Eq. (1 1 .72) If LlGo and LlHo approach each other in value, it follows that the product T LlSo -.. 0 as the temperature decreases. This could be because T is getting smaller ; however, the result was observed when the value of T was still of the order of 250 K. This strongly suggests that LlSo -.. as T -.. 0, which is the Nernst heat theorem. The validity of the third law is tested by comparing the change in entropy of a reaction computed from the third-law entropies with the entropy change computed from equi librium measurements. Discrepancies appear whenever one of the substances in the reaction does not follow the third law. A few of these exceptions to the third law were described in Section 9.17.* 1 1 . 1 7 C H E M I CA L R EACTI O N S A N D T H E E N T R O PY O F T H E U N IV E R S E

LlG = LlH - T LlS, and since the pressure is constant, Qp = LlH . The heat that flows to the surroundings is Qs = - Qp == - LlH. If we suppose that Qs is transferred reversibly to the immediate surroundings at temperature T, then the entropy increase of the surroundings is LlSs = Qs /T = - LlH/T; or LlH = - T LlSs ' In view of this relation we have LlG = - T(LlSs + LlS) . The sum of the entropy changes in the system and the immediate surroundings is the entropy change in the universe ; we have the relation

A chemical reaction proceeds from some arbitrary initial state to the equilibrium state. If the initial state has the properties T, p, G 1 , HI' and S 1 , and the equilibrium state has the properties T, p, Ge , He ' Se , then the Gibbs energy change in the reaction is LlG = Ge - G 1 ; the enthalpy change is LlH = He - HI' and the entropy change of the system is LlS = Se - S 1 ' Since the temperature is constant, we have

LlG = - T LlSunive rse ' In this equation we see the equivalence of the two criteria for spontaneity : the Gibbs energy decrease of the system and the increase in entropy of the universe. If LlSuni ve rse is positive, then I1G is negative. Note that it is not necessary for spontaneity that the entropy

246

Systems of Va r i a b l e Composition

in any spontaneous transformation.

of the system increase and in many spontaneous reactions the entropy of the system decreases ; for example, Na + ! Cl z NaCl. The entropy of the universe must increase---+

* 1 1 . 1 8 C O U P L E D R EACTI O N S

It often happens that a reaction which would be useful to produce a desirable product has a positive value of I1G. For example, the reaction

I1G98

=

+ 1 52.3 kllmol,

would be highly desirable for producing titanium tetrachloride from the common ore TiO z . The high positive value of I1Go indicates that at equilibrium only traces of TiCl4 and 0 z are present. Increasing the temperature will improve the yield TiCl 4 but not enough to make the reaction useful. However, if this reaction is coupled with another reaction that involves a I1G more negative than - 1 52.3 kllmol, then the composite reaction can go spontaneously. If we are to pull the first reaction along, the second reaction must consume one of the products ; since TiCl 4 is the desired product, the second reaction must consume oxygen. A likely prospect for the second reaction is

I1G98The reaction scheme is coupled TiO z (S) + 2 CI 2 (g) reactions C(s) + 0 2 (g) and the overall reaction is C(s) + Ti0 2 (s) + 2 CI 2 (g)-----+

=

- 394.36 kllmol.

{

-----+

-----+

TiCI 4 (l) + z Cg), C z Cg), TiC14 (l) + CO z (g),

I1G98 I1G98 I1G98=

=

=

+ 1 52.3 kllmol, - 394.4 kllmol,

- 242. 1 kllmol.

Since the overall reaction has a highly negative I1Go, it is spontaneous. As a general rule metal oxides cannot be converted to chlorides by simple replacement ; in the presence or carbon, the chlorination proceeds easily. Coupled reactions have great importance in biological systems. Vital functions in an organism often depend on reactions which by themselves involve a positive I1G ; these reactions are coupled with the metabolic reactions, which have highly negative values of I1G. As a trivial example, the lifting of a weight by Mr. Universe is a nonspontaneous event involving an increase in Gibbs energy. The weight goes up only because that event is coupled with the metabolic processes in the body that involve decreases in Gibbs energy sufficient to more than compensate for the increase associated with the lifting of the weight.1 1 . 1 9 D E P E N D E N C E O F T H E O T H E R T H E R M O DY N A M I C F U N CT I O N S O N C O M P O S IT I O N

Having established the relation between the Gibbs energy and the composition, we can readily obtain the relation of the other functions to the composition. Considering the fundamental equation, Eq. (1 1 .7