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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
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.
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