Chapter 7 Chemical Equilibrium
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Chapter 7 Chemical Equilibrium The equilibrium composition corresponds to a minimum in Gibbs energy plotted against the extent of reaction . Standard Gibbs energy of reaction Δ r G Ø  Equilibrium constant ( K ) Reactants Products

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Chapter 7 Chemical Equilibrium

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Chapter 7 chemical equilibrium

Chapter 7 Chemical Equilibrium

The equilibrium composition corresponds toa minimum in Gibbs energy

plotted against the extent of reaction.

Standard Gibbs energyof reactionΔrGØ Equilibrium constant (K)

Reactants Products

Spontaneous chemical reactions:

For spontaneous changesat constant T and p, the reaction istowards lower value of the Gibbs energy, G.

7.1. The Gibbs energy minimum:

Whenat the equilibrium compositionof a reaction mixture, theGibbs free

energyof the reaction mixture corresponds tothe minimum.


Chapter 7 chemical equilibrium

(a). The reaction Gibbs energy:

A B

Aninfinitesimal amount dεof A turns into B,

dnA = – dε; dnB = dε

Thereaction Gibbs free energy, ΔrG = (G/ε)p,T

Δr Gsignifies a derivate, the slope of G with respect toε.

dG = µA dnA + µBdnB = – µAdε + µBdε= (µB– µA) dε

ΔrG = µB– µA

µA: chemical potential of the A species; µ = (G/n)p,T

µ = (nGm/n)P,T = Gm


Chapter 7 chemical equilibrium

ΔrG = µB– µA

Because the chemical potentialsvary with composition(µ = µ0 + RT ln p),

the slope of the plot of Gibbs energy against extent of reaction changes as

the reaction proceeds.

When, µA > µB; (ΔrG < 0)

A  B is spontaneous

When µA < µB (ΔrG > 0)

B  A is spontaneous.

When µA = µB (ΔrG = 0)

The reaction mixture isat equilibrium


Chapter 7 chemical equilibrium

A and B are not mixed together.

Not common case.

A and B are mixed together.

General case


Chapter 7 chemical equilibrium

Molecular interpretation 9.1

Without mixing:

dG = GØ(B) dε – GØ(A) dε =ΔrGØdε

pure A  pure B

Mixing:

ΔmixG = nRT (A lnA + B lnB)

For the real system:

ΔrG = ΔG (no mixing) +ΔG (mixing)

Therefore,the mixing procedurehas an important contribution to the position of the chemical equilibrium.


Chapter 7 chemical equilibrium

(b) Exergonic and endergonic reactions:

The reactions for whichΔrG < 0are calledexergonic(work-producing).

The reactions for whichΔrG > 0are calledendergonic(work-consuming).

Reactions at equilibrium arespontaneous in neither direction, they are neither exergonic nor endergonic.

In biological cells, the oxidation of

carbohydrates act as the heavy weight

that drive other reactions forwards and

results in the formation of proteins,

from amino acids, muscle contraction,

and brain activity.


Chapter 7 chemical equilibrium

7.2 The description of equilibrium

(a). Perfect gas equilibria:

At constant T, dG = V dP  Gm(pf) – Gm(pi) = RT ln(pf/pi)

µ = µØ + RT ln p

The relation between chemical potential and pressure of the perfect gas:

dG = dH – TdS – S dT = dU + pdV +V dp –TdS = (TdS – pdV) + pdV + VdP – TdS – SdT = V dP – SdT; dG = Vdp – SdT


Chapter 7 chemical equilibrium

ΔrG = µB– µA = (µBØ + RT ln pB) – (µAØ + RT ln pB)

= ΔrGØ + RT ln (pB/pA) =ΔrGØ + RTlnQ ( Q = pB/pA)

Q: is an example of a reaction quotient (反應商數)

The standard reaction Gibbs energy:ΔrGØ = GB,mØ – GA,mØ

The difference in standard molar Gibbs energies of products and reactants is equal to the difference in the standard Gibbs free energy of formation.

ΔrGØ = ΔfGØ(B) – ΔfGØ(A)

At equilibriumΔrG = 0

RT lnK = –ΔrGØ ; K = (pB/pA)equilibrium

ΔrGØ < 0 ; K > 1 ; pA < pB

ΔrGØ > 0 ; K < 1 ; pA > pB


Chapter 7 chemical equilibrium

Justification :

2A + 3B C + 2D

dnA = –2 dε; dnB = –3 dε; dnC = dε; dnD = 2dε (dnJ = vJdε)

dG = µCdnC + µDdnD– µAdnA– µBdnB

= (µC + 2µD–2µA– 3µB) dε [dG = ( vJµJ)dε]

ΔrG = (G/ε)p,T = –2µA–3µB + µC + 2µD

µJ = µJØ +RT ln aJ

Q = aC aD2/aA2aB3

At equilibrium ΔrG = 0

K = (aC aD2/aA2aB3)equilibrium ; K = ( aJvJ )equilibrium

J


Chapter 7 chemical equilibrium

Justification 7.1:

The dependence of the Gibbs energy on the reaction quotient

dG = µJdnJ = ( vJµJ)dε = (( vJµJ)) dε

 ΔrG = (G/ε)p,T =  vJµJ

µJ = µJØ +RT ln aJ

 ΔrG = ( vJµJ0) + RT  vJ lnaJ

= ΔrGo + RT ln aJvJ = ΔrGo+ RT ln  aJvJ

= ΔrGo+ RT lnQ

At equilibrium ΔrG = 0 ; Q  K

K = (aC aD2/aA2aB3)equilibrium ; K = ( aJvJ )equilibrium

J

J

J

J

J

J

J

J


Chapter 7 chemical equilibrium

(b). The general case of a reaction:

The Gibbs energy of reaction can always be written:

ΔrG = ΔrGØ + RT lnQ

With the standard reaction Gibbs energy:

ΔrGØ =  vΔfGØ - vΔfGØ

General form:ΔrGØ =  vJΔfGØ(J)

The reaction quotient Q = activities of products / activities of reactants

Q =  aJvJ, for gas,aJ= fJ/pØ ; f: effective pressure,

For pure solids and liquids, theactivity is 1, such substances make no contribution to Q.

products

reactants

J


Chapter 7 chemical equilibrium

At equilibrium constant K expressedin terms of activitiesis called athermodynamic equilibrium constant. K = ( aJvJ ) equilibrium

Because the activities aredimensionless, the thermodynamic equilibrium constant isalso dimensionless.

On the other viewpoint,

K or Q can have dimension, for exampleM-1, M-2….., atm-1, atm-2…..

But ΔrG = ΔrGØ +RTlnQ,

where, Q = (pp/pp0)/(pr/pr0) in such express is dimensionless,

Consequently,the p unit must have the same unit as that of the standard state.

For example:1.0 atm, 1.0 M


Chapter 7 chemical equilibrium

Molecular interpretation 7.2: (A  B)

Boltzmann distributiondoes not distinguish between their identities,

only their energy.

ΔrGØ = ΔrHØ – TΔrSØ ;K = exp(-ΔrHØ/RT)exp(ΔrSØ/R)

Larger entropy


Chapter 7 chemical equilibrium

The response of equilibria to the conditions:

The equilibrium constant for a reaction isunaffected by thepresence of a catalyst or an enzyme(a biological catalyst).

Catalysts increase the rate (change the pathway), at which equilibrium is attained butdo not affect its position.


Chapter 7 chemical equilibrium

The responses of equilibria to the conditions

7.3 How equilibria response to pressure:

The value of K (exp(– ΔrHØ/RT)exp(ΔrSØ/R)), isindependent of pressureat which the equilibrium is actually established.

(K/p)T = 0

However, theequilibrium composition could change with the pressure.

Case I: injecting an inert gas into a reaction vessel:

Considering the ideal gas behavior, the partial pressures of the reacting gases are unchanged. (ptotal = preactant + pproduct + pinertgas)

Thusthe presence of another gas has no effect.


Chapter 7 chemical equilibrium

Case II: the pressure of the system may be increased by confining gases to a smaller container.

For instance, A 2B

K = pB2/pApØ

Tomaintain the constant K value, the relatively steep increase of pA compared to pB will occur. Thenthe number of the A will increaseas thevolume of the container is decreased.

Le Chatelier’s principle:

A systemat equilibrium, when subject toa disturbance, responds in a way thattends to minimize the effect of the disturbance.


Chapter 7 chemical equilibrium

A 2B

Initial components: n of A (no B),

At equilibrium: A is (1 – )n and the

amount of B is (2n).

A = (1– )/(1 + ); B = 2/ (1 + )

K = pB2/pA = B2p2/Ap = 42p/(1–2)

 = 1/(1 + 4p/K)1/2

Wherep = p/pØ ; relative pressure

As p increases, the  decreases.

For N2 + 3 H2 2NH3

K = p2NH3/pN2pH23 = NH3p2/N2H2p4

= K/p2


Chapter 7 chemical equilibrium

7.4 The response of equilibria to temperature:

Le chaterlier’s principle predicts:

Exothermic reactions: increased temperature favors the reactants.

Endothermic reactions: increased temperature favors the products.

Exothermic reaction:

Reactants Products + Energy

Endothermic reaction:

Reactants + Energy Products

(a). The van’t Hoff equation:

(1). dlnK/dT = ∆HrØ/RT2; (2). dlnK/d(1/T) = –ΔrHØ/R


Chapter 7 chemical equilibrium

Justification 7.2:

lnK = –ΔrGØ/RT (ΔrGØ = ΔrHØ – TΔrSØ)

dlnK/dT = –1/R[d(ΔrGØ/T)/dT] = ΔrHØ/RT2

d(1/T)/dT = – 1/T2, so dT = –T2 d(1/T)

For a exothermic reaction: (ΔrHØ < 0)

dlnK/dT (=ΔrHØ/RT2) < 0

As the temperature raises, the K value decrease, and the equilibrium shifts away from the products.

The opposite occurs in the case of endothermic reactions.

The other equation: dlnK/d(1/T) = –ΔrHØ/R


Chapter 7 chemical equilibrium

Molecular interpretation 7.3: A  B

Endothermic reactionExothermic reaction


Chapter 7 chemical equilibrium

Example 7.3:

Ag2CO3(s) AgO2(s) + CO2(g)

The original data are

K value and Temperature.

Mathematical conversion;

– lnK vs. 1/T plot is linear.

(dlnK/d(1/T) = ΔrHØ/R)

Slope = ΔrHØ/R; Intercept = ΔrSØ/R

(b). The value of K at different temperature:

To find the value of the equilibrium constant at two different temperature:

lnK2 – lnK1 = –ΔrHØ/R(1/T2 – 1/T1)


Chapter 7 chemical equilibrium

(b). The value of K at different temperature:

To find the value of the equilibrium constant at two different temperature:

lnK2 – lnK1 = – 1/R 1/T1 ΔrHØ d(1/T)

If the ΔrHØ varies only slightly with temperature over the temperature range of

interest.

 lnK2 – lnK1 = –ΔrHØ/R(1/T2 – 1/T1)

Illustration of 7.6 N2+ 3 H2 2 NH3ΔrHØ/R = 2 ΔfHØ(NH3, g)

ln K2 = ln K1 – (– 92.9 x 103 J mol-1)/(8.314 JK-1 mol-1) (1/500K – 1/298 K)

= – 1.71  K2 = 0.18

K1 (6.1 x 105, 298 K)  K2, the decrease in equilibrium constant indicates the synthesis of ammonia is an exothermic reaction.

1/T2


Chapter 7 chemical equilibrium

The extraction of metal from their oxides:

In industrials, many metals can be obtained from oxides by reduction with C or CO:

If any of the equilibria lie to the right (K > 1)

MO(s) + C(s) M(s) + CO(g)

MO(s) + ½ C(s) M(s) + CO(g)

MO(s) + ½ CO(g) M(s) + CO2(g)

To predict these above equation, we can use ΔrGØ of the measurable reactions.

(i). M(s) + ½ O2(g) MO(s)

(ii). ½ C(s) + ½ O2(g) ½ CO2(g)

(iii). C(s) + ½ O2(g) CO(g)

(iv). CO(g) + ½ O2(g) CO2(g)


Chapter 7 chemical equilibrium

Ellingham diagram

ΔrGØ = ΔrHØ – TΔrSØ

dΔrGØ/dT = –ΔrSØ

For (i). net decrease in gas amount

ΔrSØ < 0; slope < 0

For (ii). no net decrease in gas amount

ΔrSØ = 0; slope =0

For (iii). net increase in the gas amount

ΔrSØ >0; slope >0

For (iv). net decrease in the gas amount

ΔrSØ < 0; slope < 0


Chapter 7 chemical equilibrium

MO(s) + C(s) M(s) + CO(g)

ΔrGØ = ΔrGØ(iii) – ΔrGØ(i)

MO(s) + ½ C(s) M(s) + CO(g)

ΔrGØ = ΔrGØ(ii) – ΔrGØ(i)

MO(s) + ½ CO(g) M(s) + CO2(g)

ΔrGØ = ΔrGØ(iv) – ΔrGØ(i)

For the case of ΔrGØ < 0, the line of reaction (i) lies below the line for one of

the reactions (ii) to (iv).


Chapter 7 chemical equilibrium

The response of equilibria to pH

pH = – log aH3O+ ; H3O+ : hydronium ion

At low concentration, aH3O+ = [H3O+]

(a). Acid-base equilibria in water:

For an acid species HA:

HA(aq) + H2O(l) H3O+(aq) + A-(aq);

acidity constant Ka = aH3O+aA-/aHA; pKa = –log Ka

For a base B

B(aq) + H2O(l) HB+(aq) + OH-(aq)

basicity constant Kb = aHB+aOH-/aB

For water H2O: autoprotolysis

2 H2O(l) H3O+(aq) + OH-(aq)

Kw = aH3O+aOH- (Kw = 1.008 x 10-14 at 25 oC; pKw = 14.0)

pKw = pH + pOH


Chapter 7 chemical equilibrium

HA(aq) + H2O(l) H3O+(aq) + A-(aq); Ka

A-(aq) + H2O(l) HA(aq) + OH-(aq) Kb

Kw = KaKb

(b). The pH of acid and base:

Strong acids and bases are those for which the minimum Gibbs energy of the solution lies close to the products.

In contrast, the minimum Gibbs energy of weak acids and bases lies close to the reactants.


Chapter 7 chemical equilibrium

For the weak acid, the extent of proton transfer is so small,

Ka = [H3O+]2/[HA] ( [A-] = [H3O+] and [HA]  [HA]0)

pH = ½ pKa – ½ log [HA]

For the weak base B,

pOH = ½ pKb – ½ log [B]

pH = pKW – pOH = pKw – ½ pKb + ½ log [B]

(c). Acid base titration.

Strong base (MOH), B0

At the stoichiometric point of the titration,

VAA0 = VBB0

Weak acid, VA, A0


Chapter 7 chemical equilibrium

A pH curve of the titration of a weak acid with a strongbase

IV

III

II

I


Chapter 7 chemical equilibrium

At the start of titration: (I)

pH = ½ pKa – ½ log A0

After the addition of some base: (II)

HA(aq) + OH-(aq) A- + H2O(l)

[A-] = S, S is the concentration of the salt from HA + MOH

the molar concentration of the remained HA, A’ = (A0VA/V) – S

Ka = aH3O+aA-/aHA aH3O+S/A’

pH = pKa – log A’/S (Henderson-Hasselbalch equation)

The general form: pH = pKa – log [acid]/[base]

At [acid] = [base]; pH = pKa

The pH value of the half-way to the stoichiometric point.


Chapter 7 chemical equilibrium

At the stoichiometric point : (III)

All of the HA was converted to A-

[A-]  S; Kb = aHAaOH-/aA-  [OH-]2/S

[OH-] = (SKw/Ka)1/2 ( pH = pKw – pOH)

pH = ½ pKa + ½ pKw + ½ log S

For titration of a weak base with a strong acid:

pH = ½ pKa – ½ log S

After well pat the stoichiometric point, (IV)

the pH is determined by the excess base present, [H3O+]  Kw/[OH-]

pH = pKw + log B’


Chapter 7 chemical equilibrium

pKa

(d). The ‘exact’ form of the pH curve

When assume the activities of the solutes in the solution is equal to the concentration

( aA = [A])

a ‘exact’ equation of the pH curve can be obtained.

ν= KaA0[H3O+] + (Kw – [H3O+]2)([H3O+ +Ka)/

([H3O+] + Ka) ([H3O+]2 + B0[H3O+] –Kw)

ν = VB/VA


Chapter 7 chemical equilibrium

(e) Using equilibria to stabilize pH: buffers

Buffer action: the slow variation of the pH in the vicinity of S = A’

The pH value of the buffer:

pH = pKa – log [acid]/[base]

Exp. 0.20 M KH2PO4 +

0.10 M K2HPO4,

pH = pKa(KH2PO4) – log (0.2/0.1)

pH = 7.21 – 0.301 = 6.91


Chapter 7 chemical equilibrium

(f) Using equilibria to detect changes in pH: indicators

An acid-base indicators must have a color change to detect the rapid change of pH near the stoichiometric point in a titration.

Normally, they are some large, water-soluble, weakly acidic organic molecule.

HIn(aq) + H2O(l) In-(aq) + H3O+(aq)

KIn = aH3O+aIn-/aHIn

log [HIn]/[In-]  pKIn – pH

Broadly speaking,

1. An indicator with pKIn > 7 is required for the stoichiometric point at pH > 7.

2. An indicator with pKIn < 7 is required for the stoichiometric point at pH < 7.


Chapter 7 chemical equilibrium

Equilibrium electrochemistry

The ability to make very precise measurements of currents and potential differences

(‘voltages’) means that electrochemical methods can be used to determine thermodynamic properties of reactions that may be inaccessible be other methods.

Electrochemical cell: two electrodes + electrolyte (ionic conductor)

electrode + electrolyte  electrode compartment


Chapter 7 chemical equilibrium

A galvanic cell: an electrochemical cell can produce electricity as a result of

the spontaneous reaction occurring inside it.

An electrolytic cell: electrochemical cell in which a non-spontaneous

reaction is driven by an external source of current.

7.5 Half-reactions and electrodes

Oxidation: the removal of electrons from a species.

Reduction: the addition of electrons to a species.

Redox reaction: a reaction in which there is a transfer of electrons from one

species to another.

Reducing agent (or ‘reductant’): an electron donor in a redox reaction.

Oxidizing agent (or ‘oxidant’): an electron acceptor in a redox reaction.

Half‑reaction: a conceptual reaction showing the gain of electrons.


Chapter 7 chemical equilibrium

– ΔG or E

Equilibrium

Reaction extent

In a galvanic cell, a spontaneous chemical reaction generate the push-

and-pulling powers, and the extent of power depends on how close to the

equilibrium.

At equilibrium, ΔGr = 0 = we = - nFE E = 0 (no electrical potential).


Chapter 7 chemical equilibrium

Redox couple, the reduced and oxidized species in a half‑reaction.

a couple: Ox/Red

The corresponding reduction half-reaction: Ox + v e–  Red

Illustration 7.7 Expressing a recation in terms of half-reaction

AgCl(s)  Ag+(aq) + Cl–(aq) not a redox reaction

 AgCl + e–  Ag(s) + Cl–(aq) Redox couple: AgCl/Ag, Cl–

Ag(aq) + e–  Ag(s) Redox couple: Ag+/Ag

Reaction quotient: for a half-reaction, as for the reaction quotient for the overall

reaction, but the electrons are ignored.

Illustration 7.8

O2(g) + 4 H+(aq) + 4 e–  2 H2O

reaction quotient Q = aH2O2 / (aH+4)(aO2)  p0/ (aH+4)(pO2)

aH2O = 1; the oxygen behaves as a perfect gas: aO2 pO2/p0


Chapter 7 chemical equilibrium

Anode, the electrode at which oxidation occurs.

Red1  Ox1 + v e–

Cathode, the electrode at which reduction occurs.

Red2 + v e–  Red2

In galvanic cell, the cathode has a higher

(reducing) potential than the anode: the species

undergoing reduction, Ox2, withdraws electrons

from its electrode.


Chapter 7 chemical equilibrium

7.6 Varieties of Cells:

The simplest type of cell has a single electrolyte.

Daniel cell: different electrolyte

Electrolyte concentration cell:

a galvanic cell in which the electrode

compartments are identical except for

the concentrations of the electrolytes.

Electrode concentration cell: a galvanic cell in which the electrodes themselves

have different concentrations. (exp. Gas pressure or amalgams)


Chapter 7 chemical equilibrium

(a) Liquid junction potential

Liquid junction potential, Elj, the potential difference across the interface of the two electrolytes.

Between different concentrations of HCl,

the mobile H+ diffuse into the more dilute

solution. The bulkier Cl– ions follow, but

initially do so more slowly, which results

in a potential different at the junction.

Electrolyte concentration cells always

have a liquid junction; electrode concentrations

cells do not.

The concentration of the liquid junction to the

potential by joining the electrolyte compartments through a salt bridge.


Chapter 7 chemical equilibrium

(b) Notation

Cell notation: | phase boundary; liquid junction; || an interface at which the junction potential has been eliminated.

Pt(s) | H2(g) HCl(aq) | AgCl(s) | Ag(s)

A liquid junction is denoted by

Zn(s) | ZnSO4(aq) CuSO4 | Cu(s)

 Zn(s) | ZnSO4(aq) || CuSO4 | Cu(s)

Pt(s) | H2(g) | HCl(aq, b1) || HCl(aq, b2) | H2(g) | Pt(s)


Chapter 7 chemical equilibrium

7.7 The electromotive force

Cell reaction: the reaction in the cell written on the assumption that the

right-hand electrode is the cathode.

Zn(s) | ZnSO4(aq) || CuSO4 | Cu(s)

Right-hand electrode: Cu2+(aq) + 2 e–  Cu(s)

Left-hand electrode: Zn2+(aq) + 2 e–  Zn(s)

The overall cell reaction (right-hand half reaction substrates left-hand half reaction)

 Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)

(a) The Nernst equation:

A cell in which the overall cell reaction has not reached chemical equilibrium can do electrical work as the reaction drives electrons through an external circuit.


Chapter 7 chemical equilibrium

ΔrG  we (electric work)

At constant pressure and temperature, the maximum non-expansion work (we)

dwe, max = dG For a measurable change: we, max = ΔGr

Justification:

at constant temperature: dG = dH - T dS

dG = dH – TdS = dU + pdV + Vdp = dq + dw + Vdp + pdV ;

When the change is reversible, the work is maximum (wmax)

dw = dwrev = dwe, rev – pdV; and dqrev = TdS

dG = TdS + dwe, rev –p dV + Vdp + pdV

dG = dwe,rev + Vdp, at constant p  dG = dwe,rev = dwe, max .


Chapter 7 chemical equilibrium

Cell potential: the potential difference between the two electrodes

of a galvanic cell.

Electromotive force (emf): the cell potential when it is balanced by an exactly

opposing source of potential so that the cell reaction occurs reversibly, the

composition is constant, and no current flows.

emf in terms of the reaction Gibbs energy, –vFE = ΔrG.

Faraday’s constant, F = eNA.

Justification 7.3

dG = rG d  dG = dwe (maximum non-expansion work)

dwe = rG d = (– ve NAd) E = (– vFd) E

E: the potential difference between cathode and anode.

 dwe = – vFE d = rG d


Chapter 7 chemical equilibrium

Illustration 7.9 Converting emf  rG

If rG = – 1 x 102 kJmol-1

E = –rG / v F = ( – 1 x 105 J mol-1)/ [1 x (9.6485 x 104 C mol-1)] = 1 V

1 J = 1 CV

It is known that the reaction Gibbs energy

is related to the composition of the reaction

mixtureΔrG= rG0 + RT lnQ

1/vF  E= rG0/ v F + (RT/ vF) lnQ

Eo = ΔrG0/vF : standard emf

 E = Eo – (RT/vF) ln Q (Nernst equation)

The equation for the emf of a cell

in terms of the composition.


Chapter 7 chemical equilibrium

One important application of the Nernst equation is to the determination of the pH of a solution, with a suitable choice of electrodes, of the concentration of other ions.

The standard emf can be interpreted as emf when all reactant and product in the cell reaction are in their standard sates. All activity = 1  Q = 1  lnQ = 0.

Illustration 7.10

Because RT/F = 25.7 mV at 25oC,

 E = E0 – (25.7 mV/ v) ln Q

If Q is increased by a factor of 10, then the emf decreases by 59.2 mV.

(b) Cell at equilibrium:

Suppose the reaction has reached equilibrium, then Q= K,  E = 0.

K: equilibrium constant for the cell reaction, ln K = FEo/RT


Chapter 7 chemical equilibrium

Illustration 7.11 K  Eo

The standard emf of the Daniel cell is +1.10 V, the equilibrium constant for the cell

reaction Cu2+(aq) + Zn(s) ------> Cu(s) + Zn2+(aq), for which v = 2, is K = 1.5 x 1037.

Large K  the displacement of copper ion by zinc goes virtually to completion.

1.0 V is easily measurable.  It would be impossible to measure by direct

chemical analysis.

7.8 Standard potentials

Although it is not possible to measure the contribution of

a single cell, we can define the potential of one of the

electrodes as zero and then assign values to others.

Standard hydrogen electrode (SHE):

Pt(s) | H2(g)| H+(aq) Eo = 0 at all temperature.


Chapter 7 chemical equilibrium

Standard potential of a couple, Eo,the standard emf of a cell in which a couple forms the right‑hand electrode and the standard hydrogen electrode is the left‑hand electrode.

For a special case, the measurement is made on the ‘Harned cell’

Pt(s) | H2(g)| H+(aq) | AgCl(s) | Ag(s) ½ H2(g) + AgCl(s)  HCl(aq) + Ag(s)

 E = E0(AgCl/Ag, Cl-) – (RT/ F) ln(aH+aCl-/aH21/2)

We should set aH2 = 1

 E = E0 – (RT/ F) ln(aH+aCl-)

The activities can be expressed in term of the mobility b.

aH+ =  b/b0 ; aCl- =  b/bo

 E = E0 – (RT/ F) lnb2 – (RT/ F) ln 2

 E + (2RT/ F) ln b = E0 – (2RT/ F) ln 


Chapter 7 chemical equilibrium

For a 1,1-electrolyte ln   – b1/2

 E + 2RT/F ln b = E0 + cb1/2

Illustration 7.12 Determining E0

The data of b and E were obtained.

 E + 2RT/F ln b and b1/2

Plot of E + 2RT/F ln b vs. b1/2

 the intercept at b1/2 = 0 is the value

of E0 for the silver/silver chloride electrode.


Chapter 7 chemical equilibrium

An important feature of standard emf of cells and standard potentials of electrodes

is that they are unchanged if the chemical equation for the cell reaction or a half-

reaction is multiplied by a numerical factor.

A practical consequence is that a

cell emf is independent of the physical

size of the cell.

 Cell emf is an intensive property.


Chapter 7 chemical equilibrium

The standard potentials may be combined to give values for couples that are not

listed in Table 7.2.

Example 7.4. Evaluating Eo from two others

Evaluate Eo(Cu2+, Cu+) from the potentials of Cu2+/Cu and Cu+/Cu.

(a) Cu2+(aq) + 2e–  Cu(s) E0 = + 3.40 V so rG = – 2 (0.340 V) F

(b) Cu+(aq) + e–  Cu(s) E0 = + 0.522 V so rG = – (0.522 V) F

(c) Cu2+(aq) + e–  Cu+(aq) E0 = –rG / F

Because (c) = (a) – (b)

 rGo = rG0(a) – rGo(b) = – (– 0.158 V) F  E0= + 0.158 V

The general equation:

vc E0(c) = va Eo(a) + vb Eo(b)


Chapter 7 chemical equilibrium

7.9 Applications of standard potentials

Cell emf  rG ; rH ; rS; K

(a) The electrochemical series:

Two redox couples: Ox1/Red1 and Ox2/Red2

The cell: Red1,Ox1|| Red2, Ox2 ; Eo = E20 – E10

The cell reaction: Red1 + Ox2  Ox1 + Red2

If E0 > 0 (E20 > E10),the cell reaction is spontaneous.

Red1 has a thermodynamic tendency to reduce Ox2, if E1o < E20.

More briefly: low reduces high.


Chapter 7 chemical equilibrium

Electrochemical series, the metallic elements

(and hydrogen) arranged in the order of their

reducing power as measured by their standard

potentials in aqueous solution.

A metal low in the series can reduce the ions of

with higher standard potentials.

This conclusion is qualitative.

Even for reactions that are thermodynamically

favourable, there may be kinetic factors that

result in very slow rate of reaction.


Chapter 7 chemical equilibrium

I7.2 Energy conversion in biological cells:

The whole of life’s activities depends on the coupling of exergonic and endergonic reactions, for the oxidation of food drives other reactions forward.

In biological cells, the energy released by the oxidation of food is stored in adenosine triphosphate (ATP, 1)


Chapter 7 chemical equilibrium

Hydrolysis reaction of ATP:

ATP(aq) + H2O(l)  ADP(aq) + Pi– (aq) + H3O+(aq)

Pi–: an inorganic phosphate group, such as H2PO42-.

The biological standard values for ATP hydrolysis at 37oC (blood temperature) are rG = – 31 kJ mol-1, rH = – 20 kJ mol-1 and rS = + 34 JK-1 mol-1.

The hydrolysis of ATP is therefore exergonic (rG < 0) and 31 kJ mol-1 is available for driving other reactions.

In addition, because the reaction entropy is large, the reaction Gibbs energy is sensitive to temperature. The the ATP-phosphate bond of exergonicity has been called as a ‘high-energy phosphate bond’.

The ATP acts as a phosphate donor to a number of acceptors (exp. glucose), but is recharged by more powerful phosphate donors in a number of biological processes.


Chapter 7 chemical equilibrium

Glycolysis: a partial oxidation of glucose by nicotinamide adenine dinucleotide (NAD+, 2) to pyruvate ion, CH3COCO2–, continues with the citric acid cycle, which oxidizes pyruvate to CO2, and ends with oxidative phosphorylation, which reduces O2 to H2O.

Glucose is the main source of energyduring anaerobic metabolism, a form of metabolism in which inhaled O2 does not play a role.

The citric acid cycle

ATP(aq) + H2O(l)  ADP(aq) + Pi– (aq) + H3O+(aq)


Chapter 7 chemical equilibrium

The citric acid cycle and oxidative phosphorylation are the main mechanisms for the

extraction of energy from carbohydrates during aerobic metabolism (有氧代謝).

Glycolysis (醣酵解):

Glycolysis occurs in the cytosol. At blood temperature, rG = – 147 kJ mol-1. for

the oxidation of glucose by NAD+ to pyruvate ions. The oxidation of one glucose molecule is coupled to the conversion of two ADP to two ATP molecules.

C6H12O6(aq) + 2 NAD+(aq) + 2 ADP + 2 Pi– (aq) + 2H2O(l)

 2 CH3COCO2– + 2 NADH + 2 ATP(aq) + 2 H3O+(aq)

The rGo = (– 147) – 2 (– 31) kJ mol-1 = – 85 kJ mol-1

In cells that are deprived of O2, pyruvate ion is reduced to lactate ion, CH3C(OH)CO2–, by NADH.

The condition is known as muscle fatigue results from increased concentration of lactate ion after very strenuous exercise.


Chapter 7 chemical equilibrium

The citric acid cycle:

The rGo for the combustion of

glucose is – 2880 kJ mol-1.

In the presence of O2, pyruvate

is oxidized further during the citric

acid cycle and oxidative phosphorylation,

which occur in a special compartment of

cell called the mitochondrion (粒線體).

2 CH3COCO2– + 8 NAD+(aq) + + 2 FAD(aq) + 2ADP(aq) + 2 Pi– (aq) + 8H2O(l)

 6 CO2 + 8 NADH(aq) + 4 H3O+(aq) + 2 FADH2(aq) + 2 ATP(aq)

The NADH and FADH2 go on to reduce O2 during oxidative phosphorylation, which also produces ATP.


Chapter 7 chemical equilibrium

The citric acid cycle and oxidative phosphorylation generate as many as 38 ATP molecules for each glucose molecule consumed.

Each mole of ATP extracts 31 kJ from the 2880 kJ supplied by 1 mol C6H12O6, so 1178 kJ is stored for later use. Therefore, aerobic oxidation of glucose is much more efficient than glycolysis.

In the cell, each ATP molecule can be used to drive an endergonic reaction for which rG does not exceed + 31 kJ mol-1.

The biosynthesis of sucrose from glucose and fructose can be driven by plant enzyme because the reaction is endergonic to the extent rG = + 23 kJ mol-1.

The biosynthesis of protein is strongly endergonic (rH > 0 and rS < 0). The

formation a peptide link endergonic rG = + 17 kJ mol-1 and equivalent to the consumption of three ATP molecules.

 Therefore about 12 mol of glucose molecules for 1 mol of protein molecules.


Chapter 7 chemical equilibrium

The respiratory chain:

In the exergonic oxidation of glucose 24 electrons are transferred from each C6H12O6 molecule to six O2 molecules.

C6H12O6(s) + 6 H2O(l)  6 CO2(g) + 24 H+(aq) + 24 e-

6 O2(g) + 24 H+(aq)+24e-  12 H2O(l)

The electrons do not flow directly from glucose to O2.

The glucose is oxidized to CO2by NAD+ and FAD

during glycolysis and the citric acid cycle.

C6H12O6(s)+10 NAD+ +2 FAD+ 4 ADP+4 Pi- + 2 H2O

 6 CO2+10 NADH + 2 FADH2 + 4 ATP+ 6 H+


Chapter 7 chemical equilibrium

complex I

complex II

In the respiratory chain, electrons from the powerful reducing agents NADH and

FADH2 pass through four membrane-bound protein complexes and two mobile electron carriers before reducing O2 to H2O.

The reparatory chain begins in complex I (NADH-Q oxidordedutase):

H+ + NADH + Q NAD+ + QH2 E⊕= + 0.42 V, △rG⊕ = – 81 kJ mol-1

Additional Q molecules are reduced by FADH2 in complex II (succinate-Q reductase):

FADH2 + Q FAD + QH2 E⊕= +0.015 V, △rG⊕ = – 2.9 kJ mol-1


Chapter 7 chemical equilibrium

complex III

Reduced Q migrates to complex III (Q-cytochrome c oxidoredutadse), which catalyzes the reduction of the protein cytochrome c (Cyt c). Cytocrome c contains the heme c group, the central iron ion of which can exit in oxidation states + 3 and +2.

QH2 + 2 Fe3+(Cyt c) Q + 2Fe2+(Cyt c) + 2H+

E⊕= +0.015 V, △rG ⊕ = – 30 kJ mol-1

Reduced cytochrome c carries electrons from complex III to complex IV (cytochrome c oxidase), where O2 is reduced to H2O.

2 Fe2+(Cyt c) + 2 H+ + O2complexIV 2Fe3+(Cyt c)+ H2O

E⊕ = +0.815 V, △rG⊕ = –109 kJ mol-1


Chapter 7 chemical equilibrium

Oxidative phosphorylation:

The reaction that occur in complexes I, III and IV are sufficiently exergonic to drive the synthesis of ATP in the process call oxidative phosphorylation.

ADP + Pi– + H+ ATP ; △rG⊕ = + 31 kJ mol-1

In the mitocondrion, the protein complexes

associated with the electron transport chain

span the inner membrane and phosphoryaltion

take place in the matrix.

The Gibbs energy of the reactions in complexes

I, III, and IV is first used to do the work of

moving protons across the mitchondrial

membrane.


Chapter 7 chemical equilibrium

The oxidation of NADH by Q in complex I is coupled to transfer of four protons

across the membrane. The coupling of electron transfer and proton pumping in

complexes III and IV contribute further to a gradient of proton concentration across

the membrane.

The chemiosmotic theory proposed by Peter Mitchell explains how H+–ATPases synthesize ATP from ADP. The energy stored in a transmembrane protein gradient come from two contribution.

1. The difference in activity of H+ ions results in a difference in a difference in molar Gibbs energy across the mitochrondrial membrane.

△Gm,l = Gm ,in – Gm, out = RTln aH+, in/ aH+, out

2. There is a membrane potential difference  = in – out that arises from difference in Columbic interactions on each side of the membrane.

 △Gm = RT ln aH+, in/ aH+, out + F △ φ


Chapter 7 chemical equilibrium

This equation also provides an estimate of the Gibbs energy available for phosphorylation of ADP.

After using ln [H+] = ln 10 log [H+] and substituting pH = pHin – pHout

= – log [H+]in + log [H+]out

 Gm= F  – (RT ln 10) pH

In the mitochondrion, pH  – 1.4 and   0.14 V,

 so Gm  +21.5 kJ mol-1.

Because 31 kJ mol-1 is needed for phosphorylation, we conclude that at least 2 mol H+ must flow through the membrane for the phospohorylation of 1 mol ADP.


Chapter 7 chemical equilibrium

(b) The determination of activity coefficients

Once the Eo of an electrode in a cell is known, we can use it to determine mean activity coefficient by measuring the cell emf with the ions at the concentration of interest.

ln  = (E0 – E) / (2RT/F) – ln b

(c) The determination of equilibrium constants

Eo = E0(right) – E0(left)

Because Go = – vFEo , if E0 > 0, then K > 1.

Illustration 7.14

A disproportionation is a reaction in which a species is both oxidized and reduced. Exp. 2 Cu+(aq)  Cu(s) + Cu2+(aq)

Eo = + 0.52 V – 0.16 V = + 0.36 V  ln K = 0.36 / 0.025693  K = 1.2 x 106.


Chapter 7 chemical equilibrium

(d) Species-selective electrodes

An ion-selective electrode is an electrode

that generates a potential in response to

the presence of a solution of specific ions.

Exp. Glass electrode:

sensitive to H+ activity  pH.

It is filled with a phosphate buffer containing

Cl– ions, and conveniently has E = 0 when the

external medium is at pH = 7.0.

It is necessary to calibrate the glass electrode

before use with solutions of known pH.


Chapter 7 chemical equilibrium

The membrane itself is permeable to Na+ and Li+

ions but not to H+ ions. Therefore, the potential

difference across the glass membrane must arise

by a complex indirect mechanism.

A clue to the mechanism comes form a detailed

inspection of the glass membrane, for each face is

coated with a thin layer of hydrated silica.

The H+ ions in the test solution modify this layer

to an extent that depends on their activity in the

solution, and the charge modification of the outside

layer is transmitted to the inner layer by the Na+ and

Li+ in the glass.


Chapter 7 chemical equilibrium

enzyme

Electrodes sensitive to hydrogen ions and hence to pH.

 The glass can also be made responsive to Na+, K+ and NH4+ ions by being

doped with Al2O3 and B2O3.

A simple form of gas-sensing electrode consists of a glass electrode

contained an outer sleeve filled with an aqueous solution and separated from

the test solution by a membrane that is permeable to gas. When a gas such as

H2S, or NH3 diffuses into the aqueous solution, it modifies its pH, which in turn

affects the potential of the glass electrode.

Urea, amino acid ammonia  pH modification


Chapter 7 chemical equilibrium

A more sophisticated devices are used as ion-selective electrodes that give

potential according to the presence of specific ions present in the test solution.

A porous lipophilic membrane is attached to a small reservoir of hydrophobic liquid, such as dioctylphenylphosphonate. The liquid contains

an agent, such as (RO)2PO2– that

acts as a kind of solubilizing agent

for the ions with which it can form

a complex. The complex’s ion can

migrate through the lipophilic

membrane.

 transmembrane potential .


Chapter 7 chemical equilibrium

In theory, the transmembrane potential should be determined entirely by differences in activity of the species that the electrode was designed to detect.

In practice, a small potential difference, called the asymmetry potential, is observed even when the activity of the tested species is the same on both side of the membrane.

The sources of the asymmetry potential:

(1). It is not possible to produce a membrane material that has same structure

and same chemical properties throughout.

(2). All species-selective electrodes are sensitive to more than one species.

 The potential of an electrode sensitive to species X+ that is also susceptible to interference by species Y+ is given by a modified form:

E = Eap +  (RT/F) ln (aX+ + kX,Y aY+)

Eap: asymmetry potential; kx,Y: selectivity coefficient


Chapter 7 chemical equilibrium

(e) The determination of thermodynamic functions

rGo = – vFEo

Illustration 7.15 E0 fGo

Pt(s) | H2 | H+(aq) || Ag+(aq) | Ag(s) Eo = + 0.7996 V

Ag+(aq) + ½ H2(g)  H+(aq) + Ag(s) rGo = –fGo (Ag+, aq)

with v = 1  fGo (Ag+, aq) = – (–FE0) = + 77.15 kJ mol-1

The temperature coefficient of the standard cell emf, dE0/dT, gives the standard entropy of the cell reaction. ( rG0 = – vFEo)

(G/T)p = – S  dEo/dT = rS0/vF

rH0 = rG0 + TrS0 = – vF (E0 – T dE0/dT)

The standard enthalpies of formation of H+ ions in solution fH0 (H+, aq) = 0.


Chapter 7 chemical equilibrium

Example 7.5

The standard emf of cell Pt(s) | H2 | HBr(aq) | AgBr(s) | Ag(s)

Eo/ V = 0.07131 – 4.99 x 10–4 (T/K – 298) – 3.45 x 10–6 (T/K – 298)2

Evaluate the rGo, rHo, rSo at 298 K.

At T = 298 K, E0 = + 0.07131 V, rGo = – vFE = – 6.880 kJ mol-1

dE0/dT = – 4.99 x 10–4 VK-1 – 2(3.45 x 10 –6) (T/K – 298) VK-1

At 298 K, dE/dT = – 4.99 x 10–4 VK-1

rS0 = 1 x (9.6485 x 104 C mol-1) x (– 4.99 x 10–4 VK-1) = – 48.2 J K-1 mol-1

rH0 = rG0 + TrS0 = – 21.2 kJ mol-1

One difficulty with this procedure lies in the accurate measurement of small temperature coefficient of cell potential.


Chapter 7 chemical equilibrium

Exp. 2 NO2(g) N2O4(g) Exp. CoCl2(s, blue) + 6 H2O

CoCl2 . 6H2O (Pink)


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