Chem340 Physical Chemistry for Biochemists Dr. Yoshitaka Ishii Homework 7 Due Date Mar. 9, 2011
Ch 6
P6.2-6.8, P6.10, P6.13, P6.14, P6.16-6.20. P6.24, P6.28, P6.30, P6.38
P6.2) Calculate A for the isothermal compression of 2.00 mol of an ideal gas at 298 K from an initial volume of 35.0 L to a final volume of 12.0 L. Does it matter whether the path is reversible or irreversible? dA SdT PdV At constant T, we consider the reversible process. Because A is a state function, any path whether reversible or irreversible, between the same initial and final states will give the same result. Vf
A PdV nRT ln Vi
Vf Vi
2.00 mol 8.314 J mol1K 1 298 K ln
12.0 L 5.30 103 J 35.0 L
P6.3) Calculate G for the isothermal expansion of 2.50 mol of an ideal gas at 350 K from an initial pressure of 10.5 bar to a final pressure of 0.500 bar.
dG SdT VdP At constant T, we consider the reversible process. Because G is a state function, any path between the same initial and final states will give the same result. Pf
G VdP nRT ln Pi
Pf Pi
2.50 mol 8.314 J mol1K 1 350 K ln
1
0.500 bar 22.1 103 J 10.5 bar
P6.4) A sample containing 2.50 mol of an ideal gas at 298 K is expanded from an
initial volume of 10.0 L to a final volume of 50.0 L. Calculate G and A for this process for (a) an isothermal reversible path and (b) an isothermal expansion against a constant external pressure of 0.750 bar. Explain why G and A do or do not differ from one another. a) for the isothermal reversible path Pf
G VdP nRT ln Pi
Pf Pi
nRT ln
Vi Vf
2.50 mol 8.314 J mol1K 1 298 K ln Vf
A PdV nRT ln Vi
10.0 L 9.97 103 J 50.0 L
Vf Vi
50.0 L 9.97 103J 10.0 L b) Because A and G are state functions, the answers are the same as to part a) because the systems go between the same initial and final states, T,Vi → T,Vf. 2.50 mol 8.314 J mol1K 1 298 K ln
G – A = H – U = （PV) = （nRT). Therefore, G = A for an ideal gas if T is constant. 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 Gm greater for He than for graphite?
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For a solid or liquid, Pf
G VdP V Pf Pi
Pi
Gm (C , s,100 bar) Gm (C , s,1 bar) Vm Pf Pi Gm (C , s,1 bar) 0
M
P
f
Pi
12.011 103kg 99.0 105 Pa = 52.8 J 3 2250 kg m
Treating He as an ideal gas, Pf
Gm He, g ,100 bar Gm He, g ,1 bar VdP Pi
0 RT ln
Pf Pi
1 mole 8.314 J mol1K 1 298.15 K ln
100 bar =11.4 103 J 1 bar
This result is a factor of 216 greater than that for graphite. P6.6) Assuming that H f is constant in the interval from 275 to 600 K, calculate
G for the process (H2O, g, 298 K) (H2O, g, 525 K). G f ( H2O, g, 525 K).
G f T1 1 1 G f T2 T2 H f T1 T2 T1 T1 228.6 103 J mol1 1 1 525 K 241.8 103 J mol1 298.15 K 525 K 298.15 K G f 525 K 218.5 103J mol1 P6.7) Calculate Greaction for the reaction CO(g) + 1/2 O2(g) CO2(g) at 298.15
at 650 K assuming that Hreaction is constant in the K. Calculate Greaction
temperature interval of interest. 1 Greaction 298.15 K G f CO2 , g G f CO, g G f O2 , g 2 1 1 3 3 394.4 10 J mol + 137.2 10 J mol 0 257.2 103 J mol1
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1 H reaction 298.15K H f CO2 , g H f CO, g H f O2 , g 2 1 1 3 3 393.5 10 J mol + 110.5 10 J mol 0 283.0 103 J mol1
G 1 1 T Greaction T2 T2 reaction 1 H reaction T1 T1 T2 T1 257.2 103 J mol1 1 1 Greaction 283.0 103J mol1 650 K 650 K 650 K 298.15 K 298.15 K 226.8 103 J mol 1
P6.8) Calculate Areaction and Greaction 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 products. All reactants and products are treated as ideal gases Gcombustion H combustion T Scombustion Scombustion S CO 2 , g 2 S H 2 O, l S CH 4 , g 2 S O 2 , g
213.8 J mol1K 1 + 2 70.0 J mol1K 1 186.3 J mol1K 1 2 205.2 J mol1K 1 242.9 J mol1K 1 Gcombustion 890.3 103 J mol1 298.15 K 242.7 J mol 1K 1 817.9 103J mol1
Acombustion U combustion T Scombustion H combustion PV combustion T Scombustion Gcombustion T Scombustion PV T Scombustion Gcombustion nRT
where n is the change in the number of moles of gas phase species in the reaction 817.9 103J mol 1 2 8.314J mol1K 1 298.15 K Acombustion 812.9 103 J mol1
P6.10) The standard Gibbs energy of formation Gf for carbon dioxide gas is
–394.4 kJ mol–1. Calculate the Gibbs energy of formation of carbon dioxide at its normal sea level partial pressure of 0.00031 atm.
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The Gibbs energy of formation is given by: p ΔG f p ΔG 0f p0 n R T ln 0 p 0.00031 atm 394.4 kJ mol-1 1 mol 8.314472 J K -1 mol-1 298 K ln 0.99969 atm 414.4 kJ mol -1
P6.13) Nitrogen is a vital element for all living systems; except for a few types of bacteria, blue-green algae, and some soil fungi, organisms cannot utilize N2 from the atmosphere. The formation of “fixed” nitrogen is therefore necessary to sustain life and the simplest form of fixed nitrogen is ammonia NH3. Living systems cannot fix nitrogen using the gas-phase components listed in Problem P6.12. A hypothetical ammonia synthesis by a living system might be: 1 N 2 2
g 32 H2O l NH3 aq 32 O2 g
where (aq) means the ammonia is dissolved in water. Calculate the standard free energy change for the biological synthesis of ammonia and calculate the equilibrium constant as well. Based on your answer, would the biological synthesis of ammonia
occur spontaneously? Note that Gf NH3 , aq 80.3 kJ mol 1.
3 ΔG 0reaction i0 ΔG i0 1 ΔG 0NH3 aq ΔG 0H 2O 2 i 3 1 80.3 kJ mol -1 237.1 kJ mol -1 275.35 kJ mol -1 2
The equilibrium constant is then:
ΔG 0reaction 275.35 kJ mol -1 kJ mol -1 - 49 K p Exp Exp 5.45 10 -1 -1 RT 8.314472 J K mol 298 K
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P6.14) Consider the equilibrium NO2 (g) NO(g) + ½ O2 One mole of NO2(g) is placed in a vessel and allowed to come to equilibrium at a total pressure of 1 bar. An analysis of the contents of the vessel gives the following results: T
700K
800K
P(NO)/P(NO2)
0.872
2.50
a. Calculate KP at 700 and 800 K. 1
1
1
[ NO ][O2 ] 2 [ PNO ][ PO2 ] 2 [0.872][0.872] 2 0.934 K p (700 K ) [ NO2 ] [ PNO2 ] [1] 1 2
[ NO ][O2 ] K p (800 K ) [ NO2 ]
1
[ PNO ][ PO2 ] 2 [ PNO2 ]
1
[2.5][2.5] 2 3.952 [1]
0 0 b. Calculate Greaction at 298.15K assuming that H reaction is independent of
temperature. 0 G700 RT ln K P (700 K ) -0.397 kJ mol-1
0 G700 RT ln K P (800 K ) -9.140 kJ mol-1
P6.16) Consider the reaction FeO(s) + CO(g) Fe(s) + CO2(g) for which KP is
found to have the following values: T
600oC
KP
0.9
1000oC 0.396
0 0 0 a. Calculate Greaction , S reaction , and H reaction for this reaction at 600oC. Assume
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0 that H reaction is independent of temperature.
0 G873 RT ln K P (873K ) -0.764 kJ mol-1
0 G1273 RT ln K P (1273K ) -9.804 kJ mol-1
Using the Gibbs-Helmholtz equation: G T01 T1
G T02
1 1 H T2 T1
T2
G T01 G T02 H T1 T2
1 0 0 1 1 1 G1273 1 G873 1 1273 873 1273 873 29.53kJ / mol T2 T1
G H TS H G S T H G1273 S1273 22.6 J / K 1273 H G873 S873 22.6 J / K 873 b. Calculate the mole fraction of CO2(g) present in the gas phase at 600oC.
K p (873K )
X CO2
n CO
2
ntotal
[ PCO2 ] [ PCO ]
[CO2 ] [0.9] 0.900 [CO ] [1.0]
[ 0 .9 ] 0.47 [ 0 .9 1 .0 ]
P6.17) If the reaction Fe 2 N s 3 2H 2 g
comes to
2Fe s NH 3 g
equilibrium at a total pressure of 1 bar, analysis of the gas shows that at 700 and 800 K, PNH
3
PH 2.165 and 1.083, respectively, if only H2(g) was initially present in 2
the gas phase and Fe2N(s) was in excess. a. Calculate KP at 700 and 800 K.
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b. Calculate Sreaction at 700 and 800 K and H reaction assuming that it is
independent of temperature. c. Calculate Greaction for this reaction at 298.15 K.
a) FeN2(s) + 3/2 H2(g) 2Fe(s) + NH3(g)
KP
PNH 3 / P
P
H2
/ P
3/ 2
Ptotal = 1 atm = PNH 3 PH 2 At 700 K, 1 atm = 2.165 PH 2 PH 2 = 3.165 PH 2 PH 2 = 0.316 atm,
K P (700 K) =
PNH 3 = 0.684 atm
0.684
0.316
3/ 2
=3.85
At 800 K 1 atm = 1.083 PH 2 PH 2 = 2.083 PH 2 PH 2 = 0.480 atm, K P 800 K
b)
ln
PNH 3 = 0.520 atm
0.520
0.480
3/ 2
1.56
Assume that H reaction is independent of temperature
K P 800 K H reaction 1 1 800 K 700 K K P 700 K R
K P 800 K K P 700 K 42.1 kJ mol1 1 1 800 K 700 K R ln
H reaction
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Greaction 700 K RT ln K P 700 K 7.81 kJ mol 1 Greaction 800 K RT ln K P 800 K 2.91 kJ mol 1 H reaction Greaction 700 K 48.9 J mol 1 K 1 S 700 K 700 K 1 S reaction 800 K 48.9 J mol K 1 reaction
The values of Sreaction at 700 K and 800 K are nearly the same because Greaction H reaction
c) ln K P 298.15 K ln K P 700 K = ln 3.85 +
H reaction 1 1 R 298.15 K 700 K
42.1 103 J mol1 1 1 11.1 1 1 298.15 K 700 K 8.314 J mol K
Greaction 298.15 K RT ln K P 298.15 K
= 8.314 J mol1 K 1 298.15 K 11.1= 27.5 kJ mol 1 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
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
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= 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:
ΔG den 298 K ΔH den 298 K - T S den 418000 J mol -1 1300 J mol -1 K -1 298 K 30600 J mol
1
Then ΔG reaction 303 K can be calculated using: ΔG reaction T1 1 1 ΔG reaction T2 T2 ΔH reaction T1 T2 T1
30600 J mol -1 ΔG reaction 303 K 303 K 418.00 kJ mol -1 298 K 24098.97 J mol -1
3031 K 2981 K
b) The equilibrium constant at 303 K is: G den K p 303 K Exp RT
24098.97 J mol -1 J mol -1 -5 Exp 5.97 10 -1 -1 8.314472 J K mol 298 K
c) The large positive Gden and the small equilibrium constant indicate that the protein is stable. P6.19) The melting temperature of a protein is defined as the temperature at which
the equilibrium constant for denaturation has the value K = 1. Assuming that the enthalpy of denaturation is temperature independent, use the information in Problem P6.18 to calculate the melting temperature of the protein at pH = 2. 10
We first calculate the equilibrium constant at 298 K: ΔG 0reaction 298 K K p 298 K Exp RT
30600 J mol -1 -6 Exp 4.329 10 -1 -1 8.314472 J K mol 298 K
We then solve for Tm, with lnK p Tm ln1 0 : ln K p Tm ln K p 298
Tm
ΔH 0reaction R
1 1 Tm 298.15 K
1 R ln K p Tm ln K p 298 1 0 298.15 K ΔH reaction 1
1 8.314472 J K -1 mol -1 0 ln 4.329 10 -6 418000 J mol -1 298.15 K 321.5 K
P6.20) Calculate the Gibbs energy change for the protein denaturation described in
Problem 5.45 at T = 310.K and T = 340.K. From P5.45 we have:
ΔSden 310 K 1109.4 J K -1 mol -1 ΔH den 310 K 343.9 kJ K -1 mol -1 ΔH den 340 K 640.1 kJ K -1 mol -1
ΔG den 310 K is then: ΔG den 310 K ΔH den 310 K T ΔS den 310 K
343.9 kJ mol -1 310 K 1109.4 J K -1 mol -1 0 kJ mol -1
At 340 K we obtain:
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G T02 T2 0 G340
G T01
1 1 H T1 T2 T1 1 0 1 (340) (340)(640.1) 61.945kJ / mol 310 340 310
P6.24) 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: ΔG 298 K ΔH298 K ΔS298 K T
30.5 kJ mol -1 163 kJ mol -1 298 K
0.445 kJ K
-1
mol -1
To calculate G° at T = 320 K we need to calculate H and S at T = 320 K with a constant Cp,m: ΔH320 K ΔH298 K ΔC p,m ΔT
163 kJ mol -1 8.36 kJ K -1 mol -1 22 K 346.92 kJ mol -1 T ΔS320 K ΔS298 K n ΔC p, m ln 320 T298
320 K 445 J K -1 mol -1 8.36 kJ K -1 mol -1 ln 298 K 1040.46 J K -1 mol -1
G° at T = 320 K is then: ΔG 320 K ΔH320 K T ΔS320 K
346.92 kJ mol -1 320 K 1040.46 J K -1 mol -1 13.97 kJ mol -1
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P6.28) Calculate mixture 298.15 K, 1 bar O 2
for oxygen in air, assuming that the
mole fraction of O2 in air is 0.200. For pure O2:
μ O 2 pure ΔG m O 2 T ΔSm O 2 298.15 K 205.2 J mol -1 K -1 61180.38 J mol -1
For O2 in air: μ O 2 mix μ O 2 pure R T ln x O 2
61180.38 J mol -1 8.314472 J K -1 mol -1 298.15 K ln 0.200 65.2 kJ mol -1
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. c. Calculate Gmixing if 10 mol of pure He are added to the mixture of 10 mol of H2
and 10 mol of He. 1 1 1 1 a) Gmixing 20 mol 8.314 J K 1 mol1 298 K ln ln 34.4 kJ 2 2 2 2 1 1 2 2 b) Gmixing 30 mol 8.314 J K 1 mol1 298 K ln ln 47.3 kJ 3 3 3 3 c) Gmixing Gmixing pure gases Gmixing 10 mol A + 10 mol B = 47.3 34.4 12.9 kJ
P6.38) Consider the equilibrium in the reaction 3O2 g
2O3 g , with
Hreaction 285.4 103 J mol 1 at 298 K. Assume that Hreaction is independent of
temperature.
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a. Without doing a calculation, predict whether the equilibrium position will shift toward reactants or products as the 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. Calculate Kx at 550 K and 0.500 bar. a) Since the number of moles on the product side (2) is smaller than on the reactant side (3), increasing the pressure would shift the equilibrium towards the product side. b) Since the reaction is endothermic (ΔH > 0), increasing the temperature would shift the equilibrium towards the products. c) To obtain Kp at 550 K, first Kp at 298.15 has to be calculated. This requires the calculation of ΔSreaction , and subsequently ΔG reaction at 298.15 K:
ΔSreaction 298.15 K ν i Sf,i 2 238.9 J K -1 mol -1 3 205.2 J K -1 mol -1 - 137.8 J K -1 mol -1 i
ΔG 298.15 K K p 298.15 K Exp RT
285.4 10 3 J mol -1 298.15K 137.8 J K -1 mol -1 - 58 Exp 5.9865 10 -1 -1 8.314472 J mol K 550 K
Then:
1 1 4.98 10 550 K 298.15 K
285.4 10 3 J mol -1 K p 550 K Expln298.15 K 8.314472 J mol -1 K -1
d) Kx at 550 K and 0.500 bar with = 1 is given by: p K x K p p
Q1. Prove
-Δ
0.5 bar Kp 1.0 bar
1
2 K p 2 4.98 10 35 9.96 10 35
A / T 2 that U / T and (T ) V
A / T U ( 1 / T ) V
. They
are also known as Gibbs-Helmholtz equations. Write an expression analogous to Eq. (6.37) to relate A at two temperatures. (Modified from P6.27). 14
35
A / T 1 A A S A A TS 2 U / T 2 2 2 T T T (T ) V T T V T A / T A / T T A / T T 2 U T V (1 / T ) V T V (1 / T ) V
At constant V , A U T d T T d T 1 1
T2
T2
A T2 A T1 1 1 U T1 T2 T1 T2 T1
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Q2. Figure 6.8 in p126 depicts the total Gibbs energy for a mixture of NO2(g) and N2O4(g) at a different ratio in its equilibrium between (2 -2 ξ) mole of NO2 and ξ mole of N2O4(g) at a constant pressure and temperature at 1 bar and 298 K. (a) Reproduce the plots for Gpure and Gmixture assuming that Gm0(NO2, g) = -38.45 kJmol -1 and Gm0(N2O4, g) = -79.60 kJmol -1. (b) Explain how you obtained the equations to calculate Gpure and Gmixture. (c) How much is ξ in the equilibrium state. (Modified from P6.41) (a)
-77
Gpure
-1
G(kJ mol )
-78
-79
-80
Gmixture -81 0.0
-1
=0.72,G=-81.0kJ mol 0.2
0.4
0.6
0.8
1.0
G pure (2 - 2 ) G om ( NO2 , g ) G om ( N 2 O4 , g ) (2 - 2 ) (-38.45 kJ mol -1) (-79.60 kJ mol -1) -76.90 kJ mol -1 2.70kJ mol -1 G mixture G pure ΔG mixing G pure nRT ( x NO2 ln x NO2 x N 2O4 ln x N 2O4 ) (b) G pure (2 - ) RT (
2 2 2 2 ln ln ) 2 2 2 2
-76.90 kJ mol -1 2.70kJ mol -1 8.314472 10 3 298kJ mol -1 2 2 (2 - 2 ) ln ln 2 2 (c) At the equilibrium state, Gmixture has a minimum is at = 0.72.
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Q3. Plot Gmixture(ξ) at 350K for the system described in Q2 (assume that the total pressure is 1 bar). Does T affect ξ in the equilibrium state? (Hint: first, obtain Gm0(NO2, g) and Gm0(N2O4, g) at T = 400K.) 350K o 1 1 G m ( NO2 , g ,298K ) 350 K ΔH of ( NO2 , g )( ) 298K 350 K 298K 350K 1 1 (-38.45 kJ mol -1 ) 350 K (33.2 kJ mol -1)( ) 298K 350 K 298K -50.95kJ mol -1 350K o 1 1 G om ( N 2 O4 , g ,350 K ) G m ( N 2 O4 , g ,298K ) 350 K ΔH of ( N 2 O4 , g )( ) 298K 350 K 298K 350K 1 1 (-50.95 kJ mol -1 ) 350 K (11.1 kJ mol -1)( ) 298K 350 K 298K -95.43kJ mol -1
G om ( NO2 , g ,350 K )
G mixture G pure ΔG mixing G pure nRT ( x NO2 ln x NO2 x N 2O4 ln x N 2O4 ) (2 - 2 ) G om ( NO2 , g ,350 K ) G om ( N 2 O4 , g ,350 K ) nRT ( x NO2 ln x NO2 x N 2O4 ln x N 2O4 ) (2 - 2 ) (-50.95 kJ mol -1) (-95.43 kJ mol -1) nRT ( x NO2 ln x NO2 x N 2O4 ln x N 2O4 ) -101.91 kJ mol -1 6.49kJ mol -1 2 2 2 2 ln (2 - ) RT ( ln ) 2 2 2 2 -125.94 kJ mol -1 15.3kJ mol -1 8.314472 10 3 350kJ mol -1 2 2 (2 - 2 ) ln ln 2 2
T affects ξ of this reaction, ξeq corresponds to ξ~0.16.
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