university of washington department of chemistry chemistry...
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University of Washington Department of Chemistry
Chemistry 452/456 Summer Quarter 2011
Homework Assignment #7: Due at 500 pm Wednesday 17 August. Submit scanned copy as usual.
1) The following data on the osmotic pressure of solutions of the protein bovine
serum albumin (BSA) were obtained; Concentration (gL-1) 8.95 17.69 27.28 56.20 0smotic Pressure (mmHg)
2.51 5.07 8.35 19.33
From a graph of these data, determine the molecular weight of BSA. Assume T=298K
BSA Osmotic Data
00.05
0.10.15
0.20.25
0.30.35
0.4
0 10 20 30 40 50 60
C2
Osm
otic
P/C
2
Intercept is around 0.26 mmHg-L/g
Therefore
( )( )( )( )( )2
2
1 1
2 1 1 3
0
1
8.31 298
0.26 133 /1000
71,613
CC
JK mol KRTMLim mmHg L g Pa mmHg m L
gmol
π
− −
− −
→
−
= =⋅ ⋅ ⋅
=
2) Benzene and CCl4 form a regular solution at T=298K for
6 6 40.5C H CClχ χ= = . For
this solution the excess entropy of mixing 1 10.092EMIXS JK mol− −∆ = and the
enthalpy of mixing is 1110MIXH Jmol−∆ = . Calculate 11 2212 2
ZwRT
ε εε +⎛ ⎞= −⎜ ⎟⎝ ⎠
,
dwdT
, and the activity coefficient for this solution. Assume one mole total, i.e.
6 6 40.5C H CCln n= = .
Calculate also the vapor pressure of benzene over this solution. Assume the vapor pressure of pure benzene at T=298Kis 0.1252 atm.
Solution:
( )( ) ( )( )( )
( )( )( )( )
11 2
1 11 2
1 1 1 1
1 1 1
1 1 1 1
110
0.092
4 0.092 0.368
110 0.25 298 0.368
4 110 0.25 298 0.368 550
550
MIX
EMIX
dH Jmol TdT
dS JK moldT
d JK mol JK moldT
Jmol w K JK mol
Jmol K JK mol Jmol
wRT
εχ χ ε
εχ χ
ε
ε
ε
−
− −
− − − −
− − −
− − − −
∆⎛ ⎞∆ = = ∆ −⎜ ⎟⎝ ⎠
∆∆ = = −
∆∴ = − = −
∴ = − −
∆ = − − =
∆∴ = =
( )( )( )( )
( ) ( )( )
2 0.22 0.25
0
0.228.31 298
1.06
1.06 0.5 0.1252 0.066
w real
ideal
real
Pe eP
P P atm atm
χγ
γχ
=
= = = =
∴ = = =
3) Compare the mixing of molecules of similar size to the mixing of polymers. a) Assume two small molecules of similar size for which w=2.5, form a
regular solution at a temperature of T=300K and 1 0.50χ = . Calculate
MIXMIX
AAN
∆∆ = . Calculate also MIXS∆ and determine if the mixing is
driven by the enthalpy change, the entropy change or both. Calculate the activity coefficient.
Solution: ( ) ( )
( )
( )( ) ( )( )( )( )
1 1 1 11 1 2 2
1 1 2 2 1 2
1 1 1 1
1 1 1
ln ln 8.31 ln 0.5 5.76
ln ln
5.76 300 2.5 8.31 300 0.25
1728 1558 170
MIX
MIXMIX
S R JK mol JK mol
AA RT wRTN
JK mol K JK mol K
Jmol Jmol Jmol
χ χ χ χ
χ χ χ χ χ χ
− − − −
− − − −
− − −
∆ = − + = − =
∆∆ = = + +
= − +
= − + = −
The entropy of mixing is positive. The MIXT S− ∆ term is larger than the enthalpy term so 0.MIXA∆ < b) Try to mix a polymer and a solvent for which w=2.5, T=300K with the
volume fraction 1 0.50φ = . Assume the polymer has length, N=10,000.
Calculate MIXMIX
AAM
∆∆ = . Calculate also MIXS∆ and determine if the
mixing is driven by the enthalpy change, the entropy change or both. Calculate the polymer activity coefficient and the solvent activity coefficient γ. In each case are the activity coefficients dominated by interaction energy effects (i.e. by w) or by the difference in molecular sizes?
Solution:
( )
( )( ) ( )( )( )( )
1 1 1 1
1 1 1 1
1 1
ln ln
0.58.31 0.5ln 0.5 ln .5 2.8810000
ln ln
2.88 300 2.5 8.31 300 0.25
864 1558 694
pMIX s s p
pMIXMIX s s p s p
S RN
JK mol JK mol
AA RT wM N
JK mol K JK mol K
Jmol Jmol Jmol
φφ φ φ
φφ φ φ φ φ
− − − −
− − − −
− − −
⎛ ⎞∆ = − +⎜ ⎟
⎝ ⎠⎛ ⎞= − + ≈⎜ ⎟⎝ ⎠⎛ ⎞∆
∆ = = + +⎜ ⎟⎝ ⎠
= − +
= − + = 1
c) Assume two polymers for which w=2.5 are similarly mixed at T=300K so
that the volume fraction is 1 0.50φ = . Assume the polymers have lengths,
N1=10,000 and N2=50000. Calculate MIXMIX
AAM
∆∆ = . Calculate also
MIXS∆ and determine if the mixing is driven by the enthalpy change, the entropy change or both. Calculate the polymer activity coefficients and the solvent activity coefficient γ. In each case are the activity coefficients dominated by interaction energy effects (i.e. by w) or by the difference in molecular sizes?
( )
( )( )( )( )
( )( ) ( )
1 21 2
1 2
1 1
1 1 4 1 1
1 21 2 1 2
1 2
4 1 1
ln ln
0.5 0.58.31 ln 0.5 ln .550000 10000
8.31 0.693 0.5 0.00012 3.46 10
ln ln
3.46 10 300 2.5
MIX
MIXMIX
S RN N
JK mol
JK mol JK mol
AA RT wM N N
JK mol K
φ φφ φ
φ φφ φ φφ
− −
− − − − −
− − −
⎛ ⎞∆ = − +⎜ ⎟
⎝ ⎠⎛ ⎞= − +⎜ ⎟⎝ ⎠
= − − = ×
⎛ ⎞∆∆ = = + +⎜ ⎟
⎝ ⎠
= − × + ( )( )( )1 1
1 1 1
8.31 300 0.25
0.104 1558 1558
JK mol K
Jmol Jmol Jmol
− −
− − −= − + ≈
d) Based on your results for parts a-c, in which system are the components most soluable? In which system are the components least soluable? Explain.
Small molecules are most easily mixed because the entropy change is large and positive and dominates the enthalpic change…which is positive. But the entropy change is much reduced when a polymer mixes with a small molecular solvent or with another polymer. In the case of two polymers the entropy change is dominated by the enthalpy change. Result: polymers are not very miscible.
4) The standard free energy of reaction is related to the standard cell potential by the equation 0G n∆ = − ℑЄ0, where ℑ is Faraday’s constant and Є0 is the standard cell potential.
a) Prove that the standard entropy change is given by 0S nT∂
∆ = ℑ∂
Є0, at
constant pressure.
b) Prove also that 0H n∆ = − ℑ Є0 n TT∂
+ ℑ∂
Є0
c) For the redox reaction ( ) ( ) ( ) ( )2 22 2 2Ag s Hg Cl s AgCl s Hg s+ → + , the standard cell potential as a function of temperature is
T(K) 291 298 303 311 Є0(mV) 43.0 45.4 47.1 50.1
• Write out the oxidation and reduction half reactions • Using the data in the table calculate ∆G0, ∆H0, and ∆S0.
d) Prove that the standard entropy change is given by 0S nT∂
∆ = ℑ∂
Є0, at
constant pressure. Solution:
G n E H T SG E Hn S ST T T
ES nT
∆ = − ℑ = ∆ − ∆
∂∆ ∂∆ ∂∆∴ = − ℑ = −∆ ≈ −∆
∂ ∂ ∂∂∆
∴∆ = ℑ∂
e) Prove also that 0H n∆ = − ℑ Є0 n TT∂
+ ℑ∂
Є0
Solution:
G n E H T SEH G T S n E n TT
∆ = − ℑ = ∆ − ∆
∂∆∴∆ = ∆ + ∆ = − ℑ + ℑ
∂
f) For the redox reaction ( ) ( ) ( ) ( )2 22 2 2Ag s Hg Cl s AgCl s Hg s+ → + , the standard cell potential as a function of temperature is
T(K) 291 298 303 311 Є0(mV) 43.0 45.4 47.1 50.1
• Write out the oxidation and reduction half reactions Solution:
( ) ( )2
2 2 2 2
2 2 2
Ag s Cl AgCl s e
Hg e Hg
− −
+ −
+ → +
+ →
• Using the data in the table calculate ∆G0, ∆H0, and ∆S0. Make a plot of E0 versus T.
Solution:
0.0420.0430.0440.0450.0460.0470.0480.049
0.050.051
290 295 300 305 310 315
T(K)
Cel
l Pot
entia
l (V)
The plot is virtually a straight line with slope 4 13.55 10E VKT
− −∂∆= ×
∂
Assume T=298K. Then ( )( )( )
( )( )( )( )( )
10
1 4 1 1
1 1
2 96485 0.0454 8761
2 96485 3.55 10 68.5
8761 298 68.5 8761 20413 29174
G n E Cmol V J
ES n Cmol VK JKT
H G T S J K JK mol J J J
−
− − − −
− −
∆ = − ℑ∆ = − =
∂∆∆ = ℑ = × =
∂∆ = ∆ + ∆ = + = + =
•
5) Nicotine adenine dinucleotide (NAD) is cellular redox reagent. The reduced form of NAD is abbreviated NADH and the oxidized form is NAD+. In the cell, molecular oxygen O2 is reduced by NADH according to :
12 22NADH H O NAD H O+ ++ + → + .
For this reaction at T=298K ∆G0=-259.83kJ/mole. Assume [NADH]=0.035M, [NAD+]=0.004M, pH=4.8, and PO2=0.05 bars. Assume the standard oxygen pressure is 1 bar. Note the standard Gibbs energy change assumes a standard H+ concentration of 1M.
a) Calculate the standard cell potential Є0 assuming the standard state for H+ is [H+]0=1M. Repeat the calculation assuming [H+]0=10-7 M.
b) Calculate the cell voltage Є assuming the concentrations given above. Will the cell voltage depend upon the standard state definition for H+? Explain.
c) Calculate the standard cell potential Є0 assuming the standard state for H+
is [H+]0=1M. Repeat the calculation assuming [H+]0=10-7 M. Solution:
∆G = −nℑ∆E → ∆E = −
∆Gnℑ
= −−259.83kJmol -1( )1( ) 96485Cmol−1( )= 2.69V
To use the alternative standard state notation to get the cell voltage we need to know the definition of the standard free energy in the new convention. Recall:
K =
CNAD+
C0NAD+
⎛⎝⎜
⎞⎠⎟
CNADH
C0NADH( ) C
H+
CH+0
⎛⎝⎜
⎞⎠⎟
PO2
PO20
⎛⎝⎜
⎞⎠⎟
1/ 2 =
CNAD+
1M( )CNADH
1M( ) CH+
CH+0
⎛⎝⎜
⎞⎠⎟
PO2
1bar( )1/ 2 =C
NAD+
CNADH
CH+
CH+0
⎛⎝⎜
⎞⎠⎟
PO2
1/ 2
=C
NAD+
CNADH
CH+
CH+0
⎛⎝⎜
⎞⎠⎟
PO2
1/ 2=
CNAD+
CNADHCH + PO2
1/ 2 CH +0
Note if C
H +0 = 1M then
K =
CNAD+
CNADHCH + PO2
1/ 2 . We assume this definition of K in what
follows…
∆G′ = −RT ln ′K = −RT lnC
NAD+
CNADHCH + PO2
1/ 2 CH +0 = −RT ln
CNAD+
CNADHCH + PO2
1/ 2
⎛
⎝⎜⎜
⎞
⎠⎟⎟− RT lnC
H +0
= −RT lnC
NAD+
CNADHCH + PO2
1/ 2
⎛
⎝⎜⎜
⎞
⎠⎟⎟− RT lnC
H +0 = −RT ln K − RT ln10−7 = ∆G − RT ln10−7
= ∆G + 7RT ln10 = ∆G + 7 2.3( )RT = −259.83kJmol−1 + 16.1( ) 8.31JK −1mol−1( ) 298K( )= −219.96kJmol−1
∴∆E′ = −
∆G′
nℑ= −
−219.96kJmol -1( )1( ) 96485Cmol−1( )= 2.28V
d) Calculate the cell voltage Є assuming the concentrations given above.
Will the cell voltage depend upon the standard state definition for H+? Explain.
Solution:
The cell voltages in the two conventions are obtained from lnRTE E Qn
∆ = ∆ +ℑ
and
lnRTE E Qn
′′ ′∆ = ∆ −ℑ
So we have to calculate the two reaction quotients:
Q =
CNAD+
C0NAD+
⎛⎝⎜
⎞⎠⎟
CNADH
C0NADH( ) C
H+
CH+0
⎛⎝⎜
⎞⎠⎟
PO2
PO20
⎛⎝⎜
⎞⎠⎟
1/ 2 =0.004
1( )0.035
1( ) 10−4.8
1( ) 0.051( )1/ 2 =
0.0040.035( )1.58×10−5( )0.224( )
= 3.23×104
′Q =
CNAD+
C0NAD+
⎛⎝⎜
⎞⎠⎟
CNADH
C0NADH( ) C
H+
CH+0
⎛⎝⎜
⎞⎠⎟
PO2
PO20
⎛⎝⎜
⎞⎠⎟
1/ 2 =0.004
1( )0.035
1( ) 10−4.8
10−7( ) 0.051( )1/ 2 =
0.0040.035( )1.58×102( )0.224( )
= 3.23×10−3 = Q ×10−7
In the ‘1M” convention:
( )( )( )( ) ( )
1 14
1
8.31 298ln 2.69 ln 3.23 10
1 96485
2.69 0.27 2.42
JK mol KRTE E Q Vn Cmol
V V V
− −
−∆ = ∆ − = − ×
ℑ
= − =
In the “10-7M” convention: ( )( )
( )( ) ( )1 1
31
8.31 298ln 2.28 ln 3.23 10
1 96485
2.28 0.16 2.42
JK mol KRTE E Q Vn Cmol
V V V
− −−
−′′ ′∆ = ∆ + = − ×
ℑ
= + =
Therefore E E′∆ = ∆ . The cell potential is independent of the standard state definitions
6) A protein complex called the sodium/potassium pump uses the free energy of hydrolysis of ATP to pump sodium ions Na+ out of the cell and potassium ions K+ into the cell. The net reaction for active transport of sodium and potassium ions is thought to be:
3
2
3
2
Na inside
K outsideATP ADP phosphate
Na outside
K inside
+
+
+
+
+
UV|W|+ → + + +
RS|T|
b g
b g
b g
b g
The diagram below shows the concentrations of sodium and potassium ions inside and outside a cell. The electrical potential E inside and outside the cell is also given.
a) Calculate the change in the electrochemical potential involved in transporting 1 mole of sodium ion out of the cell. Assume the activity coefficients of sodium ion inside and outside the cell are unity. Assume the temperature is 310K.
b) Calculate the free energy change involved in transporting 1 mole of potassium ion
into the cell. Assume the activity coefficients of potassium ion inside and outside the cell are unity. Assume the temperature is 310K.
c) Calculate the total free energy change involved in transporting 3 moles of sodium ion
out of the cell and two moles of potassium into the cell at T=310K. Assume, as in parts a and b, that all activity coefficients are unity.
d) The standard free energy change for the hydrolysis of ATP i.e.
ATP ADP phosphate→←
+FHG
IKJ at 310K is ∆G kJ mole0 313= − . / . If the total
concentration of inorganic phosphate is 0.01M, calculate the ratio of ATP to ADP (i.e. in the reaction quotient Q) which will furnish the work required to accomplish the transport described in part c.
Solution:
( ) ( )
( )( ) ( )( )
( )( ) ( )( )
ln
1408.31 / 310 ln 96485 / 0 ( 0.07 )101408.31 / 310 ln 96485 / 0.0710
6798 6754 13552
outNa in out in out out in
in
NaG G G n RT z
Na
J mol K K C mole V
J mol K K C mole V
J J J
µ ψ ψ+
+
→ +
⎛ ⎞⎡ ⎤⎣ ⎦⎜ ⎟∆ = − = ∆ = + ℑ −⎜ ⎟⎡ ⎤⎣ ⎦⎝ ⎠
⎛ ⎞= ⋅ + − −⎜ ⎟⎝ ⎠⎛ ⎞= ⋅ +⎜ ⎟⎝ ⎠
= + =
e) Calculate the free energy change involved in transporting 1 mole of potassium ion into the cell. Assume the activity coefficients of potassium ion inside and outside the cell are unity. Assume the temperature is 310K.
Solution:
∆GK + out→in( ) = Gin − Gout = RT lnK +⎡⎣ ⎤⎦in
K +⎡⎣ ⎤⎦out
⎛
⎝⎜
⎞
⎠⎟ + zℑ ψ in −ψ out( )
= 8.31J / mol ⋅K( ) 310K( )ln 1005
⎛⎝⎜
⎞⎠⎟+ 96485C / mole( ) −0.70V( )= 1000J / mole
f) Calculate the total free energy change involved in transporting 3 moles of
sodium ion out of the cell and two moles of potassium into the cell at T=310K. Assume, as in parts a and b, that all activity coefficients are unity.
Solution: ∆Gtotal = 3∆G Na+ out( )+ 2∆G K + in( )= 3( ) 13,600J / mole( )+ 2 1000J / mole( )= 42,800J
g) The standard free energy change for the hydrolysis of ATP i.e.
ATP →←
ADP + phosphate⎛
⎝⎜⎞
⎠⎟ at 310K is ∆G kJ mole0 313= − . / . If the
total concentration of inorganic phosphate is 0.01M, calculate the ratio of ATP to ADP (i.e. in the reaction quotient Q) which will furnish the work required to accomplish the transport described in part c.
Solution: ∆Gtotal = ∆Gtotal
transport + ∆GATP→ADP ≤ 0 ∆Gtotal
transport + ∆GATP→ADP = 42,600J + ∆GATP→ADP0 + RT lnQ ≤ 0
∆Gtotaltransport + ∆GATP→ADP = 42,600J − 31,300J + 8.31J / K( ) 310K( )ln Pi[ ] ADP[ ]
ATP[ ]⎛
⎝⎜⎞
⎠⎟≤ 0
∆Gtotaltransport + ∆GATP→ADP = 11,500J + 2576J( )ln Pi[ ] ADP[ ]
ATP[ ]⎛
⎝⎜⎞
⎠⎟≤ 0
lnPi[ ] ADP[ ]
ATP[ ]⎛
⎝⎜⎞
⎠⎟≤ −4.464 ⇒
Pi[ ] ADP[ ]ATP[ ]
≤ e−4.464 = 0.01152
Pi[ ] ADP[ ]ATP[ ]
≤ 0.01152 ⇒0.01( ) ADP[ ]
ATP[ ]≤ 0.01152 ⇒
ADP[ ]ATP[ ]
≤ 1.152
Therefore, ATP[ ]ADP[ ]
≤ 0.868 .
7) The data below are for the binding of oxygen to the protein squid hemacyanin. The
percent saturation is the parameter 100% 100%fNν× = × .
PO2 (mmHg) Percent saturation 1.13 0.30 7.72 1.92 31.71 8.37 100.5 32.9 136.7 55.7 203.2 73.4 327.0 83.4 566.9 89.4 736.7 91.3 Solution:
logv
1− v versus log PO2
log v/n-v
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
0 0.5 1 1.5 2 2.5 3
log PO2
log
vbar
/1-v
bar
a) From the plot, determine whether the binding is cooperative or independent. Explain.
Solution: The slope =1.3 so the binding is not independent.
b) Estimate the number of oxygen molecules that can be attached to a single
hemacyanin molecule. Solution: The slope of this ‘line’ is 1.3, therefore N ≥ 2. 8) From an isothermal titration calorimetry study, a protein P was found to have four binding sites for ligand L (i.e. N=4). It was also found that the equilibrium constant for ligand binding was Kb=5.00x108 and the binding enthalpy was 175bH kJmol−∆ = − .
a) Determine the fraction of ligand bound and determine the heat of binding qb in a 100mL solution where the protein concentration is cP=0.0001M, and the total ligand concentration is cL(total)=0.0005M.
Solution: ( )b b Lq V H c bound= ∆ .
So we have to determine ( ) ( ) ( )L L Lc bound c total c free= − where
cL free( )=− NKbcP − KbcL total( )+1( )± NKbcP − KbcL total( )+1( )2 + 4KbcL total( )⎛
⎝⎞⎠
1/ 2
2Kb
NKbcP − KbcL total( )+1= 4( ) 5×108( )10−4 M( )− 5×108( )5×10−4 M( )+1≈ −5×104 M
∴cL free( )=5×104 M ± 25×108 M 2 + 4 5×108( )5×10−4 M( )( )1/ 2
2 5×108( )
=5×104 M ± 25×108 M 2 +100 ×104( )1/ 2
109 ≈5×104 M ± 5×104 M
109 = 10−4 M
Therefore
cL bound( )= cL total( )− cL free( )= 0.0005M − 0.0001M = 0.0004
∴qb =V∆HbcL bound( )= 0.1L( ) −75kJmol−1( )0.0004M( )= −0.003kJ
80% of the ligand is bound.
b) What is the heat of binding when the protein sites are fully saturated with
ligand? Based on this number calculate the percent saturation.
Solution:
qb =V∆HbNcP = 0.1L( ) −75kJmol−1( ) 4( ) 0.0001M( )= 0.003kJ . The protein is fully saturated (100%).