Problem Set

Assigned October 18, 2013 – Due Friday, October 25, 2013

Please show all work for credit

To hand in:

Thermodynamic Relations

1. The shells of marine organisms contain CaCO3 largely in the crystalline form known as calcite.

There is a second crystalline form of CaCO3 known as aragonite.

(a) Based on the thermodynamic and physical properties given for these two crystalline forms,

would you expect calcite in nature to convert spontaneously to aragonite given sufficient time?

Justify your answer.

(b) Will the conversion proposed in part (a) be favored or opposed by increasing the pressure?

Explain.

(c) What pressure should be just sufficient to make this conversion spontaneous at 25°C?

(d) Will increasing the temperature favor the conversion? Explain.

2. The temperature of a typical laboratory freezer unit is -20°C. If liquid water in a completely

filled, closed container is placed in the freezer, estimate the maximum pressure developed in the

container at equilibrium. The enthalpy of fusion of water may be taken as 333.4 kJ kg-1

,

independent of temperature and pressure, and the densities of ice and liquid water at -20°C are

0.9172 and 1.00 g cm-3

, respectively.

Free energy of mixing processes

3. Atkins Ch. 5: Ex. 5.2 (b)

4. Atkins Ch. 5: Ex. 5.9 (b)

5. Engel – P 6.29

A sample containing 2.25mol of He (1bar, 298K) is mixed with 3.00mol of Ne (1 bar, 298K)

and 1.75mol of Ar(1 bar, 298K). Calculate Gmixing and Smixing.

1 1

3

1 1

ln

2.25 2.25 3.00 3.00 1.75 1.757.00 mol 8.314 J mol K 298.15 K ln ln ln

7.00 7.00 7.00 7.00 7.00 7.00

18.6 10 J

ln

2.25 2.25 3.007.00 mol 8.314 J mol K ln

7.00 7.00 7.

mixing i i

i

mixing i i

i

G nRT x x

S nR x x

1

3.00 1.75 1.75ln ln

00 7.00 7.00 7.00

62.5 J K

Equilibrium

6. Atkins Ch. 6: Ex. 6.2 (b)

7. Atkins Ch. 6: Ex. 6.3 (b)

8. Engle – P.6.18 – Protein denaturing

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

native(folded) ⇌ denatured(unfolded)

For 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.

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?

a) We first need to calculate Gden at 298 K:

1

-1-1-1

dendenden

mol J 06003

K 298K mol J 1300mol J 418000S T - K 298 ΔHK 298 ΔG

Then K 303 ΔG reaction can be calculated using:

12

reaction

1

1reaction22reaction

T

1

T

1ΔH

T

TΔGTT ΔG

1-

1--1

reaction

mol J 24098.97

K 298

1

K 303

1mol kJ 418.00

K 298

mol J 30600K 303K 303 ΔG

b) The equilibrium constant at 303 K is:

5-

1-1-

1-1-

den

p

105.97 K 298mol K J 8.314472

mol J mol J 24098.97Exp

T R

GExpK 303 K

c) The large positive Gden and the small equilibrium constant indicate that the protein is stable.

9. Engle - P.6.25 – Pentene isomers

At 25°C, values for the formation enthalpy and Gibbs energy and log10 KP for the

formation reactions of the various isomers of C5H10 in the gas phase are given by

the following table:

Substance H f

kJ mol1

G f

kJ mol1 log10KP

A = 1-pentene –20.920 78.605 –13.7704

B = cis-2-pentene –28.075 71.852 –12.5874

C = trans-2-pentene –31.757 69.350 –12.1495

D = 2-methyl-1-butene –36.317 64.890 –11.3680

E = 3-methyl-1-butene –28.953 74.785 –13.1017

F = 2-methyl-2-butene –42.551 59.693 –10.4572

G = cyclopentane –77.24 38.62 –6.7643

Consider the equilibrium A⇌B⇌C⇌D⇌E⇌F⇌G, which might be established using a

suitable catalyst.

a. Calculate the mole ratios A/G, B/G, C/G, D/G, E/G, and F/G present at

equilibrium at 25°C.

b. Do the ratios of part (a) depend on the total pressure?

c. Calculate the mole percentages of the various species in the equilibrium mixture.

a)

Reaction = expproduct reactant

P

G GK

RT

⇌ / 15.27 / B

B A B A

A

xK P P

x

⇌

/ 2.745 / CC B C B

B

xK P P

x

⇌

/ 6.050 / DD C D C

C

xK P P

x

⇌

/ 0.01843 / EE D E D

D

xK P P

x

⇌

/ 442 / FF E F E

E

xK P P

x

⇌ / 4939 / G

F G G F

F

xK P P

x

Because = 0 for each reaction, Kx = KP

412.025 10

4939

F

G

x

x

7

5

6

6

8

4.581 10

2.486 10

4.109 10

1.497 10

9.803 10

E E F

G F G

D D E F

G E F G

C C D E F

G D E F G

CB B D F F

G C D E F G

CA A B D E F

G B C D E F G

x x x

x x x

x x x x

x x x x

x x x x x

x x x x x

xx x x x x

x x x x x x

xx x x x x x

x x x x x x x

b) The ratios do not depend on pressure because for each reaction is zero.

c) xG + xF + xE + xD + xC + xB + xA = 1. Because xG is much larger than all the other

mole fractures, xG 1.

xF = 2.025 10–4

, mol % = 2.025 10–2

%

xE = 4.581 10–7

, mol % = 4.581 10–5

%

xD = 2.486 10–5

, mol % = 2.486 10–3

%

xC = 4.109 10–6

, mol % = 4.109 10–4

%

xB = 1.497 10–6

, mol % = 1.497 10–4

%

xA = 9.803 10–8

, mol % = 9.803 10–6

%

10. Engle – P.6.34 – Gas reaction

Calculate KP at 475 K for the reaction NO(g) + 1/2 O2(g) NO2(g) assuming that

Hreaction

is constant over the interval from 298 to 600 K. Do you expect KP to

increase or decrease as the temperature is increased to 550 K?

2

3 3 3 3 3 3

NO , NO,

33.2 10 J mol 91.3 10 J mol 58.1 10 J mol

reaction f fH H g H g

2

3 3 3 3 3 3

2 NO , NO,

51.3 10 J mol 87.6 10 J mol 36.3 10 J mol

reaction f fG G g G g

3 1 3 1

1 1 1 1

298.15 K 1 1ln

298.15 K 298.15 K

36.3 10 J mol 58.1 10 J mol 1 1ln 475 K

8.314 J K mol 298.15 K 8.314J K mol 475 K 298.15 K

ln 475 K 5.9175

475 K 371

reaction reactionP f

f

P

P

P

G HK T

R R T

K

K

K

Because reactionH < 0, KP decreases as T increases.

11. Engle – P.6.40 – Calcium carbonate decomposition

Ca(HCO3)2(s) decomposes at elevated temperatures according to the stoichiometric equation

Ca(HCO3)2(s) CaCO3(s) + H2O(g) + CO2(g).

a. If pure Ca(HCO3)2(s) is put into a sealed vessel, the air is pumped out, and the vessel and its

contents are heated, the total pressure is 0.115 bar. Determine KP under these conditions.

b. If the vessel initially also contains 0.225 bar H2O(g), what is the partial pressure of

CO2(g) at equilibrium?

a) If pure Ca(HCO3)2(s) is put into a sealed vessel and heated, the total pressure is 0.115 bar.

Determine KP under these conditions.

Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g)

Partial pressure at P P

equilibrium, i iP x P

The total pressure is made up of equal partial pressures of H2O(g) and CO2(g).

2 2 2

2 2

30.1153.31 10

2

H O CO H O

P

P P PK

P P P

b) If one of the products is originally present

Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g)

Partial pressure at

equilibrium, i iP x P P + Pi P

2 2

2

2

30.225 3.31 10

0.0139; 0.0139 bar

H O CO iP

CO

CO

P P P P P P PK

P P P P P P

PPP

P P

12. Oxides of sulfur are important in atmospheric pollution, arising particularly from

burning coal. Use the thermodynamic data at 25°C given in the table below to answer

the following questions.

(a) In air, the oxidation of SO2 can occur: 1/2O2(g)+SO2(g)=SO3(g). Calculate the

standard Gibbs free energy for this reaction at 25°C.

(b) Find the equilibrium ratio of partial pressures of SO3(g) to SO2(g) in air at 25°C;

the partial pressure of O2(g) is 0.21 atm.

(c) SO3(g) can react with H2O(g) to form sulfuric acid, H2SO4(g). Air that is in

equilibrium with liquid water at 25°C has a partial pressure of H2O(g) of 0.031 atm.

Find the equilibrium ratio of partial pressures of H2SO4(g) to SO3(g) in air at 25°C.

13. What is the pressure of CO(g) in equilibrium with the CO2(g) and O2(g) in the

atmosphere at 25°C? The partial pressure of O2(g) is 0.2 atm, and the partial pressure of

CO2(g) is 3 atm. CO is extremely poisonous because it forms a very strong

complex with hemoglobin. Should you worry?

Extras – practice for exam, do not hand in

Thermodynamic Relations

14. Atkins – P.3.30 – Joule coefficient derivation

Since U is a state function, so we can write it as U=U(V,T), thus:

(

) (

) (

)

And (

) , so:

(

)

Now dU=TdS-pdV, plut it into the right hand side:

(

)

Meanwhile,

(

) , a conclusion from PS.7 Prob. 5, therefore:

15. Atkins – P.3.24 – Ideal gas derivation

Plugging dU=TdS-pdV in the numerators will show the answer.

16. Atkins - E.3.22(b) - Free energy of isothermal process

17. Engle – P.6.23 – Glucose breakdown

Under anaerobic conditions, glucose is broken down in muscle tissue to form lactic

acid according to the reaction: C6H12O6 2 CH3CHOHCOOH. Thermodynamic data

at T = 298 K for glucose and lactic acid are given below.

H f

kJ mol1

CP ,m J K

1mol

1 S°(J K–1

mol–1

)

Glucose –1273.1 219.2 209.2

Lactic Acid –673.6 127.6 192.1

Calculate G° at T = 298 K and T = 310. K. Assume all heat capacities are constant from T = 298

K to T = 310.K.

G° at T = 298 K is given by:

-1 -1

-1 -1 -1 -1

-1 -1 -1 -1

ΔG 298 K ΔH 298 K T ΔS 298 K

2 673.6 kJ mol 1 1273.1 kJ mol

298 K 2 192.1 J K mol 1 209.2 J K mol

74.1 kJ mol 298 K 175.0 J K mol 126.3 kJ mol

To calculate G° at T = 310 K we need to calculate H and S at T = 310 K:

p,m

-1 -1 -1 -1 -1

-1

ΔH 310 K ΔH 298 K ΔC ΔT

74.1 kJ mol 2 127.6 J K mol 1 219.2 J K mol 12 K

73.668 kJ mol

1-1-

1-1-1-1-1-1-

298

310

mp,

mol K J 76.421

K 298

K 310lnmol K J 19.221mol K J 27.612mol K J 75.01

T

Tln ΔCn K 298ΔSK 310ΔS

G° at T = 310 K is the

-1 -1 -1 -1

ΔG 310 K ΔH 310 K T ΔS 310 K

73.668 kJ mol 310 K 176.42 J K mol 128.4 kJ mol

18. Engle – P.6.24 – Protein denaturing*

At T = 298 K and pH=3 chymotrypsinogen denatures with G° = 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 S0at T = 298 K:

1-1-1-1-

mol K kJ .4450 K 982

mol kJ 631 mol kJ 0.53

T

K 298ΔH K 298ΔGK 298ΔS

To calculate G° at T = 320 K we need to calculate H and S at T = 320 K with a constant

Cp,m:

1-1-1-1-

mp,

mol kJ 46.923 K 22mol K kJ .368mol kJ 631

ΔTΔCK 298ΔHK 320ΔH

1-1-

1-1-1-1-

298

320

mp,

mol K J 040.461

K 298

K 320lnmol K kJ .368mol K J 454

T

Tln ΔCn K 298ΔSK 320ΔS

G° at T = 320 K is then:

1-1-1-1- mol kJ 3.971 mol K J 040.461K 203mol kJ 46.923

K 320ΔS TK 320ΔHK 320ΔG

Free energy of mixing processes

19. Engle – P.6.31 – Gas mixture*

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles of Xe is prepared at a

pressure of 1 bar and a temperature of 298 K. The total number of moles in the mixture is

three times that of Ar. Write an expression for Gmixing in terms of x. At what value of x

does Gmixing have its minimum value? Calculate Gmixing for this value of x. If the number of moles of Ne is x, the number of moles of Xe is y = 8 – x.

ln

4 4 8 8ln ln ln

12 12 12 12 12 12

1 12 1 8 8 12ln ln 1 0

12 12 12 12 12 12 8

1 1 8ln 1 ln 1 ln 0

12 12 12 12 12 8

1; 48

4 4ln

12 12

mixing i i

i

mixing

mixing

G nRT x x

x x x xnRT

d G x x x xnRT

dx x x

x x nRT xnRT

x

xx

x

G nRT

1 1 3

4 4 4 4ln ln

12 12 12 12

112 mol × 8.314 J mol K 298.15 K ln 32.7 10 J

3

20. Atkins Ch. 5: Ex. 5.10 (b)*

Equilibrium

21. Engle – P.6.26 – Temperature dependent enthalpy error*

In this problem, you calculate the error in assuming that Hreaction

is independent of T for a

specific reaction. The following data are given at 25°C:

CuO(s) Cu(s) O2(g)

H f

kJ mol1 –157

G f

kJ mol1 –130

CP,m J K

1mol

1 42.3 24.4 29.4

a. From Equation (6.71),

d ln KP

KP

T0

KP T f

1

R

Hreaction

T2T0

Tf

dT

To a good approximation, we can assume that the heat capacities are independent of

temperature over a limited range in temperature, giving

Hreaction

T Hreaction T0 CP T T0 where

CP viCP,m i i . By

integrating Equation (6.71), show that

ln KP T ln K P T0 Hreaction

T0 R

1

T

1

T0

T0 CP

R

1

T

1

T0

CP

Rln

T

T0

b. Using the result from part (a), calculate the equilibrium pressure of oxygen over copper and

CuO(s) at 1200 K. How is this value related to KP for the reaction ( ) ⇌ ( )

( )?

c. What value would you obtain if you assumed that Hreaction

were constant at its value for

298.15 K up to 1200 K?

0

0

0 0 0

2

0 2

0

2 2

0

0

1ln

1

1 1l

f f

f

f f f

T T

reactionP

T T

T

reaction P

P f P

T

T T T

reaction P P

T T T

reaction P

f

Ha) d K dT

R T

H C T T ln K T ln K T dT

R T

H T C C TdT dT dT =

R T R T R T

H T C =

R T T R

0

0 0

1 1n

f P

f

T C T

T R T T

b) 2CuO(s) 2Cu(s) + O2(g)

3 1 3 1

0

, , 2 ,

1 1

1 1

2 157 10 J mol 314 10 J mol

2 Cu, O , 2 CuO,

2 24.4 29.4 2 42.3 J K mol

6.4 J K mol

reaction

P P m P m P m

H T

C C s C g C s

3 1

1 1

3 1 1 1

1 1 1 1

1 1

2 130 10 J molln 1200 K

8.314 J mol K 298.15 K

2 157 10 J mol 1 1 6.4 J K mol 1200 K ln

8.314 J K mol 1200 K 298.15 K 8.314 J K mol 298.15 K

6.4 J K mol 298.15 K

8

PK

2

2

1 1

5

5

1 1

.314 J mol K 1200 K 298.15 K

ln 1200 K 10.1818

1200 K 3.78 10

3.78 10 bar

P

O

P

O

K

PK

P

P

c) This is equivalent to setting CP = 0. Neglecting the last two terms in the calculation above

gives ln KP = –9.6884 and 2OP = 6.20 10

–5 bar.