1

Physical Chemistry I for BiochemistsChem340Chem340

Lecture 23 (3/07/11)

Yoshitaka IshiiCh 6.10-6.12 HW7

Ch 7.1-7.4

• P6.5) The pressure dependence of G is quite different for gases and condensed phases. Calculate Gm(C, solid, graphite, 100 bar, 298.15 K) and Gm(He, g, 100 bar, 298.15 K) relative to their standard state values. By what factor is the change in G greater for HeBy what factor is the change in Gm greater for He than for graphite?

'),(),(0

000 P

P mmmm dPVTPGTPGG

)()( PPVgraphiteG [Q1] )()( 0PPVgraphiteG mm

)/ln(')/(')( 000PPRTdPPRTdPVHeG

P

P

P

P mm

[Q1]

[Q2]

2

• P6.8) Calculate and for the reaction CH4(g) + 2O2(g) CO2(g) + 2H2O(l) at 298 K from the combustion enthalpy of methane and the entropies of the reactants and productsentropies of the reactants and products.

• Greaction0 is easy to obtain.

• Greaction0 = Hreaction

0 -TSreaction0

The question is how to get Areaction0

A 0 G 0 (PV)• Areaction0 = Greaction

0 - (PV)

= Greaction0 - (ngas)RT Q. How do you get this (PV)?

We had a similar question between U and H.

6.7 Expressing Chemical Equilibrium in Terms of the i

• mAA + mBB + mCC .. mMM + mNN + mOO• We write the reaction in an abbreviated form as

- mAA - mBB - mCC .. + mMM + mNN + mOO =0 A B C M N O

• iXi = 0 (i = -mi if Xi is a reactant; i = mi if Xi is a product).

• dG associated with change in ni is given bydG = idni (*)

Now, we introduce extent of the reaction (0≤ ≤ 1)n = n initial + dn = dni = ni + i dni = i d

ni = niinitial + i = mi(1- ) for reactant

= mi for product

when 100 % of reactants reacted.

3

HW Q2 (text P126-127)• 2NO2(g) ↔ N2O4(g)(2 - 2) mole mole

G = (2 - 2)G NO20 + G N2O4

0

Q3. Where is the equilibrium point?

Gpure (2 2)Gm,NO2 + Gm,N2O4

Gmixture = Gpure + Gmix

Gmix = nNO2RTln(xNO2)

+ nN2O4RTln(xN2O4)

2Gm,NO20

G 0Gm,N2O40

Q1. How much is Greaction0 ?

Greaction0 = - 2Gm,NO2

0 + Gm,N2O40

Q2. How much is Greaction ?

[Q1]

Greaction = Greaction0 - 2RTln(XNO2)+RTln(XN2O4)[Q2]

HW Q3 (text P126-127)• 2NO2(g) ↔ N2O4(g)(2 - 2) mole mole

G = (2 - 2)G NO2(T) + G N2O4(T)

Q1 How do you read Gm,N2O4

from the graph below?

Q2. How about Gm,NO2 ?Gpure (2 2)Gm,NO2(T) + Gm,N2O4(T)Gmixture = Gpure + Gmix

Gmix = nNO2RTln(xNO2)

+ nN2O4RTln(xN2O4)

T =350 K

-120

-115

-110

Plot @ 400K(don’t use itfor HW)

Jmol

-1)

xNO2 = nNO2/{nNO2+nN2O4}= (2-2)/(2-)

Q3. Where is the equilibrium point?

xN2O4 = nN2O4/{nNO2+nN2O4}= /(2-)

-130

-125

120

0 0.5 1

Gpure

Gmixture

G(k

J

4

6.7 Expressing Chemical Equilibrium in Terms of the i

• mAA + mBB + mCC .. mMM + mNN + mOO• We write the reaction in an abbreviated form as

- mAA - mBB - mCC .. + mMM + mNN + mOO =0 A B C M N O

• iXi = 0 (i = -mi if Xi is a reactant; i = mi if Xi is a product).

• dG associated with change in ni is given bydG = idni (*)

Now, we introduce extent of the reaction (0≤ ≤ 1)n = n initial + dn = dni = ni + i dni = i d

From (*)dG = iid = Greactiond

. .

(G/)T,P =0 System reaches equilibrium.

(G/)T,P = ii =Greaction..

• For a compound, consider a formation reaction from elements (A, B): A + 2B C

G 0 G 0 G 0 2G 0

Reaction Gibbs Energy (p127)

Gf0 = Gm,C

0 - Gm,A0 - 2Gm,B

0

• For a reaction (PA=PB=PC=PD=1bar): A +2B C + 2D• Greaction = Gm,C

0 +2Gm,D0 - Gm,A

0 - 2Gm,B0

= Gf,C0 +2 Gf,D

0 - Gf,A0 - 2Gf,B

0

In general,

• Gf0 = Gm,product

0 + i Gm,reactant i0

Greaction0 = iGf.i

0

For a pure element, Gf0 = 0. [Q1]

Greaction0 = iGm.i

0

= iμi0

5

6.8 Calculating Greaction and Introducing the Equilibrium Constant for a Mixture of Ideal Gas• 2A + B ↔3C + D Greaction = iGf.i = 3Gf.C(PC) + Gf.D(PD)

@ Partial Pressure PA, PB, ..

• mAA + mBB ↔ mCC + mDD

reaction i f.i f.C( C) f.D( D)– 2Gf.A (PA) – Gf.B (PB)

= 30C + 3RTln(PC/P0) + 0

D + RTln(PD/P0) - 20

A - 2RTln(PA/P0) - 0B - RTln(PB/P0)

= 30C + 0

D - 20A - 0

B

mBeq

mAeq

mDeq

mceq

p

PP

PP

PP

PP

K

00

00

BA

DC

mAA + mBB ↔ mCC + mDD

Greaction = iGf.i = Greaction0 + iRTln(Pi/P0)

=Greaction0 + RTlnKp

+ [Q1]

Q. What isGreaction0 ? Greaction

0 = iμ0i

6.8 Continued

Greaction = iGf.i = Greaction0 + iRTln(Pi/P0)

=G 0 + RTlnK=Greaction0 + RTlnKp

• At equilibrium, Greaction =0

RTlnKp = -Greaction0 mB

eqmA

eq

mDeq

mceq

p

PP

PP

PP

PP

K

00

00

BA

DC

Thermodynamic Equilibrium Const

Kp = exp(- Greaction0/RT) (at 298K)

When A ↔ B Kp = PA/PB = exp(-G0/RT)

6

6.9 Calculating the Equilibrium Partial Pressure in a Mixture of Ideal Gases

• Cl2(g) ↔ 2Cl(g) For the reaction above, derive an expression for Kp in terms of n0,

, and P assuming the reaction starts from n0 mole of Cl2.

Total n0 + ξeq

Kp = exp(-Greaction0/RT)

Greaction0

= 2Gf, Cl0 - Gf, Cl2

0

6.10 Variation of Kp with T

2

000 1

RT

H

dT

TGd

RdT

RTGd

dT

Kdreactionreactionreactionp

)/()/()ln(

HTG / dT >0 & H > 0 dln(K ) >0 2T

H

T

TG

P

/

Gibbs-Helmholtz

dT

RT

HTKTKK

Tfin

Tini

reactioninipfinpp 2

0

)}(ln{)}(ln{)ln(

dT >0 & H > 0 dln(Kp) >0More product

dT >0 & H < 0 dln(Kp) < 0Less product

KTR

H

RT

H

fin

reaction

Tfin

Tini

reaction

298

1100

KTR

HKKTK

fin

reactionpfinp 298

11)}298(ln{)}(ln{

0- Greaction

0/RT0-Greaction(Tf)/RTf

7

7.10 Conformational Transitions of Biological Polymers

N ↔ DCini(1- fD) CinifD

D

D

iniN

iniD

N

D

f

f

Cf

Cf

C

CK

1

At Tm K =1 & fD = 0.5D

• HW 7 P6.18) Many biological macromolecules undergo a transition called denaturation. Denaturation is a process whereby a structured, biological active molecule, called the native form, unfolds or becomes unstructured and biologically inactive. The equilibrium is N (Native) ↔ D (Denatured)

• For a protein at pH = 2, the enthalpy change associated withFor a protein at pH 2, the enthalpy change associated with denaturation is H° = 418.0 kJ mol–1 and the entropy change is S° = 1.3 kJ K–1 mol–1.

a. Calculate the Gibbs energy change for the denaturation of the protein at pH = 2 and T = 303 K. Assume the enthalpy and entropy are temperature independent between 298.15 and 303 K.

G °(298K) = H °- TS °[Q1 ] GD (298K) = HD - TSD

b. Calculate the equilibrium constant for the denaturation of protein at pH = 2 and T = 303 K.

c. Based on your answers for parts (a) and (b), is protein structurally stable at pH = 2 and T = 303 K?

[Q1 ]

121

122

11

TTH

T

TGTTG D

D [Q2 ]

KD(T2) = exp(- GD/(RT2))

8

6.11 Equilibria Involving Ideal Gases and Solid or Liquid Phase

• We consider equilibrium in a thermal decomposition of CaCO3:

CaCO (s) CaO(s) + CO (g)• CaCO3(s) CaO(s) + CO2(g)

In this case, gas (CO2) is in equilibrium with two solid phase (CaCO3 &CaO)

Greaction = ii = 0

0 = eq(CaO,s,P) + eq(CO2,g,P) - eq(CaCO3,s,P)

(C O P) 0(C O ) (C CO P) 0(C CO )eq(CaO,s,P)~ eq0(CaO,s) ; eq(CaCO3,s,P)~eq

0(CaCO3,s)

0 = 0(CaO,s) + 0(CO2,g) - 0(CaCO3,s) + RTln(PCO2/P0)

Greaction0 = - RTln(PCO2/P0)

6.12 Expressing the Equilibrium Constant in Terms of Mole Fraction

0000 PPx

PPx

PP

PP

mDeq

mceq

mDeq

mceq

DCDC

0

0000

P

P

xx

xx

PPx

PPx

PP

PP

PP

PPK

mBmAmDmc

mBeqmAeq

mDeqmceq

mBeq

mAeq

mBeq

mAeq

p

BA

DC

BABA

0

0

P

PKK

P

PK

PX

X

BA

9

• HW7 P6.14 Consider the equilibrium

NO2(g) NO(g) +1/2O2(g)

One mole of NO2(g) is placed in a vessel and allowed to come to ilib i l f 1 b A l i f h

(1- y) y (1/2) y

equilibrium at a total pressure of 1 bar. An analysis of the contents of the vessel gives the following results:

T 700 K 800 K

PNO/PNO2 0.872 2.50

a. Calculate KP at 700 and 800 K. 2

1

OpNOp 2

1

y

y

1n

n

x

x

p

p872.0

222 NO

NO

NO

NO

NO

NO

@700K

b. Calculate G0reaction for this reaction at 298.15 K,

assuming that H0reaction is independent of temperature.

2

1

21

2

2

21

px

xx

p

Op

NOp

NOp

pNOp

pOp

pNOp

KNO

ONO2

21

2

2

p

p

HW8 P6.15• Consider the equilibrium CO(g) + H2O(g) ↔ CO2(g) + H2(g)

At 1000 K, the composition of the reaction mixture is

CO2 H2(g) CO(g) H2O(g) Substance (g)

27 1 27 1 22 9 22 9 M l %• 27.1 27.1 22.9 22.9 Mole %

a. Calculate KP and at 1000 K.

b. Given the answer to part (a), use the of the reaction species to calculate at 298.15 K. Assume that is independent of temperature.

922922

1.271.270

22

mBmAmDmc

HCOp P

PxxK [Q1] [Q2]

H f

Greaction

9.229.220

2

OHCO

p Pxx

)1000K

1-ΔH G

1000K

298.15K

ΔHΔHΔHΔHΔH

oreactionf,

oreaction

oCf,

oHf,

oCf,

oHf,

oreactionf, 22

KKKG

TKRTTG

oreaction

OOO

preaction

298

11000298

2

()()(

)(ln)(

KTR

HKKTK

fin

reactionpfinp 298

11)}298(ln{)}(ln{

0

10

6.7 Expressing Chemical Equilibrium in Terms of the i

• mAA + mBB + mCC .. mMM + mNN + mOO• We write the reaction in an abbreviated form asWe write the reaction in an abbreviated form as

- mAA - mBB - mCC .. + mMM + mNN + mOO =0

• iXi = 0 (i = -mi if Xi is a reactant; i = mi if Xi is a product).

Q11 From the following data at 298.15 K and 1 bar, choose the closest value to the standard formation enthalpy for CO(g). Reactions Reaction enthalpy (kJmol-1) Fe2O3(s) + 3C(graphite) 2Fe(s) + 3CO(g) 493 FeO(s) + C(graphite) Fe(s) + CO(g) 156 CO2(g) CO(g) + 1/2O2(g) 283

266

A

B

C

Fe(s)+ 1/2O2(g) FeO(s) -266 D

2A Fe(s) + 3A CO(g) – A Fe2O3(s) -3A C(gra) =0B Fe(s) + B CO(g) – B FeO(s) -B C(gra) =0

C CO(g) + C/2 O2(g) – C CO2(g) =0D FeO(s) – D Fe(s) – D/2 O2(g) =0

(2A+ B-D)Fe(s) + (3A+B +C) CO(g) –AFe2O3 -(3A+B)C(gra)(-B+D) FeO(s) –C CO2(g) +(C/2 – D/2)O2(g) =0

Hf(CO(g)) is calculated for the following reaction:C(grap) + 1/2O2(g)CO(g)CO(g) – C(grap) –1/2O2(g) = 0

2A+ B-D =0; -A =0;-B + D =0; -C = 0 A = 0, C = 0, B =DFor CO, (3A + B +C) =1 B =1 & D = B = 1 .

11

7.1 What Determines the Relative Stability of the Solid, Liquid, and Gas Phase

• Phase: Solid Liquid Gas

Phase refers to “a form of matter that isPhase refers to a form of matter that is uniform with respect to chemical composition and the state of aggregation”

Q. What determines most stable phase?

solid(T, P), liquid(T, P), gas(T, P)[Answer here?]solid( , ), liquid( , ), gas( , )[ ]

How does in three phases depends on T?

d(T, P) = -SmdT + VmdP = -SmdT

0 < S (T) < S (T) < S (T)

(T, P) = (0, P) - SmT

0 < Sm, solid(T) < Sm, liquid(T) < Sm, gas(T)

(T) for 3 phases

entia

l ()

Q1. Where is the melting i t?

Assuming Sm is constant,

Che

mic

al p

otepoint?

Q2. Which line represents solid(T)?

Q3. How do you explain the phase transition from solid to liquid using the figure?

12

(T) plot when P is increased- How are Tm and Tb affected by P?

VmGas > Vm

Liquid > VmSolid Vm

Gas > VmSolid > Vm

Liquid

(T, P + P) ~ {(0, P) + Vm P} – SmT

Sublimation and Triple Point

Sublimation TempSolid Gas

Triple Point

Solid, Gas, and Liquid Coexist

Triple Point Temp

13

7.2 The Pressure-Temperature Phase Diagram

P-T Phase Diagram (VmSolid < Vm

Liquid)Isobar T Increases • S L G

L-G coexists above critical point(T ≥Tc, P ≥ Pc) Supercritical Fluid

Water• S L G(P higher than triple point Pressure)b) S G (P lower than triple point P)

Isotherm; P Increasesc) G L S (T higher than triple point T)

• P7.3) Within what range can you restrict the values of P and T if the following information is known about CO2? Use Figure 7.8 to answer this problem.

a. As the temperature is increased, the solid is first converted to the liquid and subsequently to the gaseous state.

b. As the pressure on a cylinder containing pure CO2 is increased from 65 to 80 atm, no interface delineating liquid and gaseous phases is observed.

c. Solid, liquid, and gas

phases coexist at equilibrium.

b80

aa. 5.11 atm < P < 73.75 atm

14

7.4 Providing a Theoretical Basis for the P-T Phase Diagram

• When two phase and are in equilibrium at a constant P and T,

(P, T) = (P, T) When (P,T) is changed by (dP,dT) while keeping the

equilibrium,(P, T) +d = (P, T)+ d

Thus, d = dd d

-SmdT +VmdP = -SmdT +VmdP 0 = (Sm -Sm)dT - (Vm - Vm)dP = SmdT - VmdP

Clapeyron equation

m

m

V

S

dT

dP

dP/dT for fusion and vaporization

1122

KJmolSdP

fusion

m

Average of Sm and Vm for Ag, AgCl, Ca, CaCl2, KCl, Na, NaCl

For Fusion

116

136

551055

104

KbarPaK

molmVdT fusionfusion m

m

.

~

F V i ti Average of S and V in Table 7.3

Negative sign only for H2O(dT/dP)fusion ~ 0.05K bar-1

For Vaporization

1213

132

11

1051005

102

95

KbarPaK

molm

KJmol

V

S

dT

dPvap

m

vapm

vap

.

~

Average of Sm and Vm in Table 7.3

(dT/dP)vap ~ 20K bar-1

15

7.5 Using the Clapeyron Equation to Calculate Vapor Pressure as a Function of T

2RT

HP

TV

H

TV

H

V

S

dT

dP vapm

vapm

vap

vapm

vap

vapm

For Vaporization

2RTTVTVVdT gasmvap

mvap

mvap ,

dTTnR

HdP

P

vapm

2

11

dTH

dPvapm

TfinPfin 11

Clausius-Clapeyron equation

dTTR

dPP

m

TiniPini2

inifin

vapm

ini

fin

TTR

H

P

P 11ln

• P6.17) If the reaction Fe2N(s) +3/2H2(g) 2Fe(s)+NH3(g)comes to equilibrium at a total pressure of 1 bar, analysis of the gas shows that at 700 and 800 K, PNH3/PH2 = 2.165 and 1.083, respectively, if only H2(g) was initially present in the gas phase and Fe2N(s) was in excess. a. Calculate KPat 700 and 800 K. b. Calculate Sreaction

0 at 700 and 800 K reactionand Hreaction

0 assuming that it is independent of temperature. c. Calculate Greaction

0 for this reaction at 298.15 K.

• Fe2N(s) + 3/2H2(g) 2Fe(s) + NH3(g)Nini nini

Nini - ξ nini – 3/2ξ 2ξ ξ

a K = exp(-G0/RT) = (P /P )/(P /P )3/2 (*)a. Kp = exp(-G /RT) = (PNH3/P0)/(PH2/P0) ( )PNH3 = xNH3 P0; PH2 = (1-xNH3) P0

PNH3/PH2 = 2.165 = xNH3 /(1-xNH3) xNH3 @700Kb. Ln{Kp(800K)} - Kn{Kp(700K)} = -(H/R)(1/800K – 1/700K) G = H - TS. From (*) G = -RTln(Kp)

16

• P6.24) At T = 298 K and pH=3 chymotrypsinogen denatures with

G° = 30 5 kJ mol–1 H° = 163 kJ mol–1 and C = 8 36 kJG = 30.5 kJ mol 1, H = 163 kJ mol 1, and CP,m = 8.36 kJ

K–1 mol–1. Determine G° for the denaturation of

chymotrypsinogen at T = 320. K and pH=3. Assume CP,m is

constant between T = 298 K and T = 320. K.

• First, we calculate S0 at T = 298 K:

0 0 0Use S0 = (G0 - H0)/T

H(320K) = CP,m T + H0

S(320K) = + S0 = S0 + Cp,mln(Tfin/Tini)Tfin

Tini

dTT

Cp

• P6.30) You have containers of pure H2 and He at 298 K and 1 atm pressure. Calculate Gmixing relative to the unmixed gases of

a. a mixture of 10 mol of H2 and 10 mol of He.

b. a mixture of 10 mol of H2 and 20 mol of He.

Gmixing = nARTln(xA) + nBRTln(xB)

c. Calculate Gmixing if 10 mol of pure He are added to the mixture of 10 mol of H2 and 10 mol of He.

Gmixing = Gb - Ga

17

• P6.38) Consider the equilibrium in the reaction

3O2 ↔ 2O3 with at 298 K. Assume that is independent of temperature.

• a. Without doing a calculation, predict whether the equilibrium position will shift toward reactants or products as

Hreaction 285.4 103 J mol1

Hreaction

q p pthe pressure is increased.

• b. Without doing a calculation, predict whether the equilibrium position will shift toward reactants or products as the temperature is increased.

• c. Calculate KP at 550 K.

d C l l t K t 550 K d 0 500 b• d. Calculate Kx at 550 K and 0.500 bar.

• Kp(298K) = exp(-G/RT)

• Use ln(Kp(550K)) = ln(K(298K)) –H{1/T2-1/T1}/R

Kp = Kx(P/P0)v Kx = (P/P0) -vKp