An Introduction to Thermodynamics phần 8 pdf

and therefore The last equation can be integrated, assuming ∆H is constant with temperature, to giveo In these equations K is the equilibrium constant at the temperature T , K the equili

( ) ( )

( )

torr 0.119 atm 10

x 56 1

0125 0

ln 840

, 10

4

-2 / 1

2 / 1 2

2

2

2 2

=

=

=

=

−

=

=

∆

=

O

O

o

O Ag

O Ag

P

P K

K RT G

a

a a K

( )( )

10 -2

10

x 78 1

ln 55900

2

=

−

=

=

∆

K

K RT G

a

a a K o CaF

F Ca

This is a quadratic equation, with the solution x = 0.0428 atm The equilibrium condition may

now be more explicitly interpreted for this example If 2 atm of ammonia comes to equilibrium with its decomposition products, the equilibrium state will be 1.9572 atm of NH , together with3 0.0214 atm of N and 0.0642 atm of H Slight further decomposition, at these concentrations,2 2 would cause no change in free energy; there is no tendency for further decomposition because the system is now at equilibrium An equilibrium constant for a reaction involving gases, in which

pressures appear as an approximation for activities, is often written K (and, because pressures P are substituted for dimensionless activities, K may have apparent dimensions of powers of P

pressure)

Consider now the problem of finding the equilibrium pressure of oxygen above Ag O at2

25 C.o

Ag O -6 2 Ag + ½ O2 2

Thus silver oxide, placed in an evacuated system, should decompose until the oxygen pressure reaches 0.119 torr In an atmosphere containing 20% oxygen the reaction would be driven in the other direction, so that silver should react with oxygen to form silver oxide The slow rate of this reaction (in either direction) is attested by the durability of silver goods An equilibrium constant

for dissociation of a compound is often written Kdiss or, when the dissociation is into ions, Kionization For the solubility of CaF in water at 25 C,2 o

CaF -6 Ca + 2 F2 ++

-The equilibrium constant for dissolution of a slightly soluble salt in water is called the solubility

product constant, or K sp

The activity of solid CaF is 1 Assume that activities of the ions are numerically equal to2

concentrations From the dissociation equation we find that if the concentration of Ca is s, the++

concentration of F will be 2s, so long as there is no other source of these ions than CaF

-2 Inserting these symbols into the equation above gives

M s

s

10

x 6 7

1 0

10

x 78 1 4

4

10 3

=

=

M

s

10

x 5 3

0.5

x 1 0

10

x 78 1

9

10

=

=

1.78 x 10 = s(2s)

s = 3.5 x 10 M-4

This is the solubility of CaF , because each mole of CaF going into solution gives one mole of2 2

Ca ions Introduction of activity coefficients would probably change the calculated value by++ something on the order of 20%, which is often negligible in determining solubilities (Reported experimental values differ by much more than this for some compounds, but such accuracies are entirely adequate for some applications.)

To find the solubility of CaF in 0.5 M NaCl the deviation of activity coefficients from 12

should not be neglected The equations may be written

K = 1.78 x 10 = (a ++)(a -) = (γ ++)(c ++)(γ -) (c -)-10 2 2 2

= Q (γ)Q (c)

Experimental measurements indicate that Q (γ), the correction factor, is on the order of 0.1 for such a solution Solving for the concentrations as before gives

The increase in solubility when the other ions (Na and Cl ) are added is called the salt effect The+

-added salt lowers the activity coefficients of all ions present

The solubility of CaF in 0.5 M NaF will be smaller because of the common-ion effect This2

is apparent when we substitute numbers into the equations obtained above

K = 1.78 x 10 = -10 Q (γ)(c ++)(c -)2

Now the concentration of fluoride is no longer equal to twice the concentration of calcium ion,

but is 0.5 + 2s when the Ca has the concentration s The correction term may be assumed the++

same as for NaCl Then

1.78 x 10 = 0.1s(0.5 + 2s)-10 2

Assume first that 2s is negligible with respect to 0.5 Then

This answer confirms the assumption that taking s negligible with respect to 0.5 gives a

self-consistent, and therefore mathematically valid, solution to the problem

An orderly procedure in working problems of chemical equilibrium will greatly decrease the

2 PH

3 2

3

2

a

a a

K eq = P H

2 PH

3 H

3

2

a a

opportunity for mistakes The recommended steps are as follows:

1 Write the chemical reaction, even though it is a familiar one

2 Write out the equilibrium constant expression, Q (a), in terms of activities and set this equal to K eq

3 Determine the equilibrium constant for the reaction as written, either from the constant

given or from standard-state free energies Note that reversing a reaction inverts the equilibrium constant, and that multiplication of the equation by an integer raises the equilibrium constant to that power

4 Make the appropriate substitutions of symbols (that is: concentrations, pressures, mole fractions, or unity) for each of the activities Include activity coefficient if necessary (either numerically or symbolically)

5 Determine, from information given and the stoichiometry of the chemical reaction, what is known about concentrations or pressures

6 Solve the algebraic equation

7 Examine the solution to be sure that the answer found is the answer to the question asked and is a reasonable value in terms of the physical problem as stated and the approximations made

in the solution of the problem

Consider, for example, the decomposition of the gas phosphine, PH (∆G = 18.24 kJ/mol at3 o

25 C), to give white phosphorus (solid) and hydrogen gas If 5 atm of PH in a closed vesselo

3 comes to equilibrium with its decomposition products, what will be the final pressure of PH ? 3 The specific steps of the solution, as described above, are:

1 2 PH3W 2 P + 3 H2

2

3 ∆G = - 36.5 kJ = - 8.314 x 298 ln Ko

K = 2.51 x 106

4 a = 1; 2.51 x 10 = P 6

5 2 PH -6 2 P + 3 H3 2

5 - 2 x 3 x x = pressure, atm

2.51 x 10 = (3x) /(5 - 2x)6 3 2

6 Find the solution to the equation 2.51 x 10 (5 - 2x) = 27x 6 2 3

Because K is large, the reaction must be shifted well to the right, so x will be nearly equal to 5/2,eq and 5 - 2x will be very sensitive to x Therefore, as a first approximation, let x = 5/2 on the right-hand side and find (5 - 2x):

( ) ( )

2 2

1 1

ln

RT

G S

T RT

G RT

S

dT

T d R

G dT

G d RT RT

G dT

d dT

K d

o o

o o

o o

o

∆ +

∆

=

∆ +

∆

=

∆

−

∆

−

=











∆

−

(18)

ln 2 RT H dT K d = ∆ o (19)

ln

2 1 1

2

T RT

T H K

K = ∆ o∆

(5 - 2x) = 27(5/2) /2.51 x 10 = 1.68 x 10

x = 2.49

This calculated value is in excellent agreement with the value assumed, so the calculation is self-consistent

7 The final pressure of PH is 5 - 2x = 0.013 atm and of H , 3x = 7.48 atm This is a3 2

reasonable answer for the problem given, because ∆G (PH ) = 18.24 J/mol indicates that PH iso

quite unstable

Temperature Dependence of Equilibrium Constants

The equilibrium constant does vary with temperature From equation 17, ln K = - ∆G /RT,o

so we can find the temperature dependence Take the derivative with respect to temperature (holding the pressure constant at its standard value), employing equation 30 of Chapter 2

and therefore

The last equation can be integrated, assuming ∆H is constant with temperature, to giveo

In these equations K is the equilibrium constant at the temperature T , K the equilibrium2 2 1

constant at T , ∆H is the enthalpy change for the reaction with all reactants and products in their1 o

standard states, and ∆T = ∆T -∆T 2 1

The similarity of equation 19 to the Clausius-Clapeyron equation (equation 6, Chapter 3) is not accidental Equations for solubility, for reaction rate constants, and for other quantities, have equivalent forms This form is to be expected because all of these physical properties depend on the system surmounting an energy barrier (corresponding to the heat of reaction, the heat of vaporization, the heat of solution, or some other such energy, or enthalpy, term) and therefore

they depend, in a fundamental way, on the Boltzmann distribution law, which is an exponential

expression relating numbers of molecules, energy states, and temperature, and which therefore gives rise to a logarithmic temperature dependence

( ) (22)

ln a n

RT o

Q F

−

=E E

(23)

ln K

n

RT o

F

Electrochemistry

The best method of directly measuring the free-energy change of a chemical reaction is often

to carry out the reaction in an electrochemical cell The free-energy change, at constant

temperature and pressure, is (eqn 26, Chapter 2)

(T,P) dG = w + PdV = w ’ rev rev

or

∆G = W ’ (20) rev where W ’ is the electrical work Electrical power, P, is I R = EI and electrical work is power rev 2 multiplied by time, or potential multiplied by charge:

W ’ = Pt = EI t = Eq rev

The potential, E, is in volts, the current I in amperes, the time t in seconds, the resistance, R, in ohms, the charge q, in coulombs, and power, P, in watts, and the work w’ in joules A convenient unit of charge is a mole of electrons (6 x 10 electrons), called a faraday One faraday, F, is23

96,487 coulomb If n mol of electrons (or n “equivalents”) are transferred at a potential

difference E the work done is W ’ = Eq = E(nF) and the free-energy change is rev

∆G = - nFE (21)

Thus ∆G is obtained directly from E.

For example, we will find that the standard cell potential (all reactants and products present

at unit activity) for the reaction that transfers 2 electrons from copper to ferric iron (Fe+++) is 1.100 V The standard free energy (per mol Cu 6 Cu ) is therefore++

∆G = - nFE = - 2 x 96,487 x 1.100 = - 21,230 J/molo Substitution of equation 21 into equation 7 gives

-nFE = - nFE + RT ln o Q(a)

and therefore

This is now commonly called the Nernst equation Similarly, by combining equation 21 with

equation 17 we obtain

( ) Q( )c n

RT Q

n

RT o

ln ln

F F

E

( 1) ln ( )c (24)

n RT o i Q F E E = − = γ (25)

P P n T T G S       ∂ ∂ =       ∂ ∆ ∂ − = ∆ F E (26)

P T T n n











∂

∂ +

−

=

Writing

a = γ c i i i

for each of the reactants and products, the function Q factors to give

Q(a) = Q(γ)Q(c)

and thus

If each of the activity coefficients, γ , is unity, then i Q(γ) = 1, ln Q(γ) = 0, and

The potential depends on the actual activities, or concentrations, of the reactants and products

Not only ∆G and equilibrium constants can be determined electrochemically, but also ∆H and

∆S From equation 21 and equation 30 of Chapter 2,

and therefore, from ∆H = ∆G + T∆S, we obtain

HALF-REACTION POTENTIALS An electrical potential, or electrochemical potential, exists for any chemical reaction in which electrons are transferred from one substance to another Such

reactions are called oxidation-reduction reactions, or simply “redox” reactions Removal of

electrons, as in Fe 6 Fe + 3e, or 2 I 6 I + 2e, is called oxidation; addition of electrons is+3

-2

called reduction.

Electrons enter or leave a device through a conductor called an electrode The cathode

(from the Greek “down way”) is the electrode at which electrons enter any device; the anode (from the Greek “up way”) is the electrode at which electrons leave any device To push

electrons through a vacuum tube or electroplating cell, the cathode must be made negative On the other hand, the electrode at which an electrochemical cell releases electrons is called negative

(Figure 1) and this is, by definition, the anode.

It is not possible to measure half a reaction Only complete reactions (cathode reaction + anode reaction) can be measured Nevertheless, it is convenient to consider the electrochemical

potential of a reaction as a sum of the potentials of two half reactions, each defined from equation 21

in terms of the free-electron changes Let ∆G be A

the free-energy change for the anode reaction

(electrons leaving, hence oxidation), ∆G be the C

free-energy change for the cathode reaction

(electrons entering, hence reduction), and ∆G be

the free-energy change for the total reaction

Free-energy changes are additive, so

∆G = - nFE = ∆G + ∆G = - n FE - n FE A C A A C C and because n = n = n for any balanced reaction, A C

E = E + E (27) A C

Note that the potentials are independent of n and

thus independent of the amount of reaction! That

is, the potential of a half-reaction such as

½ M + e -++ 6 ½ M

is identically the same as the potential of the half-reaction

M + 2 e -6 M++

In nearly all electrochemical cells there is a small additional term that should be included on the right-hand side of equation 27, arising from the contact between dissimilar solutions Like E andA

E , these “junction potentials” cannot beC measured, nor can they be calculated by thermodynamics They can, however, be

Because of the inherent asymmetry of an

electrochemical cell, oxidation, occurring at the

anode, with release of electrons, and reduction,

occurring at the cathode, with removal of electrons

from the circuit, produces a potential difference

between anode and cathode that drives charges

around the circuit Charge is conserved There is

no “source” of charge, or “source” of current.

estimated, based on theories involving diffusion rates, and are usually sufficiently small in practice

that they can be ignored for present purposes.

Although it is not possible to measure E or E separately, it is possible to measure total cell A C

potentials Then, by choosing an arbitrary value for some half-cell potential as a reference level,

other reaction potentials can be determined relative to this arbitrary reference The

half-reaction chosen as the reference for this purpose is

2 e + 2 H -6 H+

2

for which E (the potential with all reactants and products at unit activity) is arbitrarily assignedo the value zero Then, for example, because the standard potential of the cell reaction

Zn + 2 H -6 H + Zn+ ++

2

Table 1 STANDARD ELECTRODE POTENTIALS, 25 C (STANDARD REDUCTION POTENTIALS) O

Reaction E (volt) o Reaction E (volt) o

Li + e + 6 Li -3.0401 Cu + e ++ 6 Cu + 0.153

K + e + 6 K -2.931 SO + 4 H + 2e = + 6

4

Na + e + 6 Na -2.71 H SO + H O 0.172

Mg + 2e ++ 6 Mg -2.372 AgCl + e 6 Ag + Cl - 0.22233

Al +++ + 3e 6 Al -1.662 Cu + 2e ++ 6 Cu 0.3419

Mn(OH) + 2e 2 6 Cu + e + 6 Cu 0.521

Mn + 2 OH - -1.5 I + 2e 6 2 I - 0.5355

2

Cr + 2e ++ 6 Cr -0.913 MnO + e - 6 MnO4 = 0.558

4

Zn + 2e ++ 6 Zn -0.7628 MnO + 2 H O + 3e

2 CO (g) + 2 H + 2 e2 + 6 MnO + 4 OH2 - 0.595

6 H C O (aq)2 2 4 -0.49 O + 2 H + 2e 2 + 6 H O 2 2 0.695

Fe + 2e ++ 6 Fe -0.447 Fe +++ + e 6 Fe ++ 0.771

Cr +++ + 2e 6 Cr ++ -0.407 Hg ++ + 2e 6 2 Hg 0.7973

2

Cd + 2e ++ 6 Cd -0.4030 Ag + e + 6 Ag 0.7996

AgI + e 6 Ag + I - -0.15224 Hg + 2e ++ 6 Hg ++ 0.851

2

Sn + 2e ++ 6 Sn -0.1375 2 Hg + 2e ++ 6 Hg ++ 0.920

2

Pb++ + 2e 6 Pb -0.1262 NO + 4 H + 3e 3 6

O + H O + 2e 2 2 NO + 2 H O2 0.957

6 HO + OH2- - -0.076 Cr O + 14 H + 6e 2 7= +

Cu(NH )3 4++ + 2e 6 2 Cr + 7 H O +++ 2 1.350

6 Cu + 4 NH3 -0.05 Cl + 2e 2 6 2 Cl - 1.35827

Fe +++ + 3e 6 Fe -0.037 MnO + 8 H + 5e - + 6

4

2 H + 2e + 6 H 0.0000 Mn + 4 H O ++ 1.507

AgBr + e 6 Ag + Br- 0.07133 Cr ++++ + e 6 Ce +++ 1.72

Hg Br + 2e 2 2 6 MnO + 4 H + 3e 4 6

+

2 Hg + 2 Br- 0.13923 6 MnO + 2 H O 2 2 1.679

Sn ++++ + 2 e 6 Sn ++ 0.151 H O + 2 H + 2e + 6 2 H O 1.776

has been found to be 0.763 V, it follows that the standard potential for the half reaction

Zn -6 Zn + 2 e++

An electric potential difference is the work required, per unit of charge, to move a charge 4

between two points, just as a gravitational potential difference is the work required, per unit of mass, to move a mass between two points The gravitational potential difference for a well is positive if the top is measured with respect to the bottom, or negative if the bottom is measured with respect to the top, and an electric potential that is positive if the cathode is measured with respect to the anode will be negative if the reaction is reversed (The “standard electrode

potential” includes a choice of sign in the definition, but “standard potential” simply means that the reactants and products are in their standard states.)

For more half-reaction potentials see especially W Latimer, Oxidation Potentials, 2

ed., Prentice-Hall, Englewood Cliffs, N.J., 1952

must be 0.763 V In turn, from the measured E for the reactiono

Zn + Cu -6 Cu + Zn++ ++

of 1.100 V, it follows that for

2 e + Cu -6 Cu++

the standard potential is E = 0.337 V We can proceed in this fashion to find the potential foro any half reaction on this arbitrary scale Reversal of a cell reaction changes the sign of a

potential The more positive a half-reaction potential, the greater the tendency of that half4

reaction to proceed as written (But it should be kept in mind that the reaction may be too slow

to observe or that some other spontaneous reaction may occur instead.)

Some standard potentials for reduction half-reactions are given in Table 1 According to the

Stockholm Convention, such standard reduction potentials are called standard electrode

potentials If the reaction is written in reverse order, E will have the opposite sign and is calledo

the standard oxidation potential Tabulations of half-reaction potentials are often given for

oxidation half-reactions.5

CALCULATIONS OF CELL POTENTIALS A few examples will illustrate the applications of standard potentials of half reactions Consider the cell reaction

2 Fe+++ + Cu -6 Cu + 2 Fe++ ++

This consists of the two half reactions,

2 e + 2 Fe+++ -6 2 Fe++

Cu -6 Cu + 2 e++

The standard potential for the first is taken directly from Table 1, that for the second is obtained from the table simply by changing the sign (because the reaction is the reverse of that given in the

V 130 1

1 0

01 0 ln C/equiv 96,487

x equiv/mol 2

K 298

K x J/mol 314 8 V

130 1

ln

=

⋅

−

=

−

=

+ +

Cu Zn

Zn Cu o

a a

a a n

RT

F E

E

( ) 0.05915log ( ) (28)

ln

C) (25 a log 2.306 x 96487 x V 298.15 x 314 8 ln

o

a Q n

a Q n RT

n a

Q n RT

=

=

F

( )25oC o 0.05915log ( )a (28a)

−

=E E

table) Thus

E = 0.771 + (-0.337) = 0.434 Vo

The positive cell potential indicates that the original reaction will proceed as written, with substances at unit activity

The cell reaction shown in Figure 2 is

Zn 6 Zn++ + 2e and Cu++ + 2e 6 Cu The potential, at 25 C, may be found as follows From Table 1,o

E = 0.763 + 0.337 = 1.100 Vo

assuming the solid metals, Zn and Cu, are at unit activity (pure and not seriously strained) The activities of the ions have been approximated by their concentrations

The concentration correction factor, (RT/nF) ln Q (a),

appears sufficiently often that it is helpful to carry out a partial evaluation At room temperature, 25 C, we may writeo

Equation 22 may therefore be written

The Nernst equation may also be applied to half-reaction potentials For example, the potential of the reduction half-reaction

Zn + 2 e -6 Zn++

would be