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;
17.69
27.28
56.20
Concentration (gL-1) 8.95
0smotic Pressure
2.51
5.07
8.35
19.33
(mmHg)
From a graph of these data, determine the molecular weight of BSA. Assume
T=298K
Osmotic P/C2
BSA Osmotic Data
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
10
20
30
40
50
60
C2
Intercept is around 0.26 mmHg-L/g
8.31JK −1mol −1 ) ( 298 K )
(
RT
=
M2 =
Lim π C2 ( 0.26mmHg ⋅ L ⋅ g −1 )(133Pa ⋅ mmHg −1 )( m3 /1000 L )
Therefore
C2 → 0
= 71, 613 gmol −1
2) Benzene and CCl4 form a regular solution at T=298K for χ C6 H 6 = χ CCl4 = 0.5 . For
E
= 0.092 JK −1mol −1 and the
this solution the excess entropy of mixing ∆S MIX
enthalpy of mixing is ∆H MIX = 110 Jmol −1 . Calculate w =
Z
RT
ε11 + ε 22 ⎞
⎛
⎜ ε12 −
⎟,
2 ⎠
⎝
dw
, and the activity coefficient for this solution. Assume one mole total, i.e.
dT
nC6 H 6 = nCCl4 = 0.5 .
Calculate also the vapor pressure of benzene over this solution. Assume the vapor
pressure of pure benzene at T=298Kis 0.1252 atm.
Solution:
d ∆ε ⎞
⎛
∆H MIX = 110 Jmol −1 = χ1 χ 2 ⎜ ∆ε − T
⎟
dT ⎠
⎝
d ∆ε
E
∆S MIX
= 0.092 JK −1mol −1 = − χ1 χ 2
dT
d ∆ε
∴
= −4 ( 0.092 JK −1mol −1 ) = −0.368 JK −1mol −1
dT
(
∴110 Jmol −1 = ( 0.25 ) w − ( 298 K ) ( −0.368 JK −1mol −1 )
(
)
)
∆ε = 4 110 Jmol −1 − ( 0.25 )( 298 K ) ( −0.368 JK −1mol −1 ) = 550 Jmol −1
∴w =
550
∆ε
=
= 0.22
RT ( 8.31)( 298 )
γ = e wχ = e( 0.22)( 0.25) = 1.06 =
2
Preal
Pideal
∴ Preal = γχ P 0 = (1.06 )( 0.5 )( 0.1252atm ) = 0.066atm
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 χ1 = 0.50 . Calculate
∆A
∆AMIX = MIX . Calculate also ∆S MIX and determine if the mixing is
N
driven by the enthalpy change, the entropy change or both. Calculate the
activity coefficient.
Solution:
∆S MIX = − R ( χ1 ln χ1 + χ 2 ln χ 2 ) = − ( 8.31JK −1mol −1 ) ln 0.5 = 5.76 JK −1mol −1
∆AMIX
= RT ( χ1 ln χ1 + χ 2 ln χ 2 ) + wRT χ1 χ 2
N
= ( −5.76 JK −1mol −1 ) ( 300 K ) + ( 2.5 ) ( 8.31JK −1mol −1 ) ( 300 K )( 0.25 )
∆AMIX =
= −1728 Jmol −1 + 1558 Jmol −1 = −170 Jmol −1
The entropy of mixing is positive. The −T ∆S MIX term is larger than the
enthalpy term so ∆AMIX < 0.
b) Try to mix a polymer and a solvent for which w=2.5, T=300K with the
volume fraction φ1 = 0.50 . Assume the polymer has length, N=10,000.
∆AMIX
. Calculate also ∆S MIX and determine if the
M
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:
φp
⎛
⎞
∆S MIX = − R ⎜ φs ln φs + ln φ p ⎟
N
⎝
⎠
Calculate ∆AMIX =
0.5
⎛
⎞
ln .5 ⎟ ≈ 2.88 JK −1mol −1
= − ( 8.31JK −1mol −1 ) ⎜ 0.5ln 0.5 +
10000
⎝
⎠
∆AMIX =
φp
⎛
⎞
∆AMIX
= RT ⎜ φs ln φs + ln φ p + wφsφ p ⎟
M
N
⎝
⎠
= ( −2.88 JK −1mol −1 ) ( 300 K ) + ( 2.5 ) ( 8.31JK −1mol −1 ) ( 300 K )( 0.25 )
= −864 Jmol −1 + 1558 Jmol −1 = 694 Jmol −1
c) Assume two polymers for which w=2.5 are similarly mixed at T=300K so
that the volume fraction is φ1 = 0.50 . Assume the polymers have lengths,
∆A
N1=10,000 and N2=50000. Calculate ∆AMIX = MIX . Calculate also
M
∆S MIX 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?
⎛φ
⎞
φ
∆S MIX = − R ⎜ 1 ln φ1 + 2 ln φ2 ⎟
N2
⎝ N1
⎠
0.5
⎛ 0.5
⎞
= − ( 8.31JK −1mol −1 ) ⎜
ln 0.5 +
ln .5 ⎟
10000
⎝ 50000
⎠
= − ( 8.31JK −1mol −1 ) ( −0.693)( 0.5 )( 0.00012 ) = 3.46 × 10−4 JK −1mol −1
∆AMIX =
⎛φ
⎞
∆AMIX
φ
= RT ⎜ 1 ln φ1 + 2 ln φ2 + wφ1φ2 ⎟
M
N2
⎝ N1
⎠
= ( −3.46 ×10−4 JK −1mol −1 ) ( 300 K ) + ( 2.5 ) ( 8.31JK −1mol −1 ) ( 300 K )( 0.25 )
= −0.104 Jmol −1 + 1558 Jmol −1 ≈ 1558 Jmol −1
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 ∆G 0 = − nℑ Є0, where ℑ is Faraday’s constant and Є0 is the standard cell
potential.
∂ 0
a) Prove that the standard entropy change is given by ∆S 0 = nℑ
Є , at
∂T
constant pressure.
∂ 0
Є
b) Prove also that ∆H 0 = − nℑ Є0 + nℑT
∂T
c) For the redox reaction 2 Ag ( s ) + Hg 2Cl2 ( s ) → 2 AgCl ( s ) + 2 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.
∂ 0
Є , at
d) Prove that the standard entropy change is given by ∆S 0 = nℑ
∂T
constant pressure.
Solution:
∆G = − nℑE = ∆H − T ∆S
∂∆G
∂∆E
∂∆H
= − nℑ
=
− ∆S ≈ −∆S
∂T
∂T
∂T
∂∆E
∴∆S = nℑ
∂T
∴
e) Prove also that ∆H 0 = − nℑ Є0 + nℑT
∂ 0
Є
∂T
Solution:
∆G = − nℑE = ∆H − T ∆S
∂∆E
∂T
f) For the redox reaction 2 Ag ( s ) + Hg 2Cl2 ( s ) → 2 AgCl ( s ) + 2 Hg ( s ) , the
∴∆H = ∆G + T ∆S = − nℑE + nℑT
standard cell potential as a function of temperature is
T(K)
291
298
303
0
Є (mV)
43.0
45.4
47.1
311
50.1
•
Write out the oxidation and reduction half reactions
Solution:
2 Ag ( s ) + 2Cl − → 2 AgCl ( s ) + 2e−
2 Hg 2+ + 2e− → 2 Hg
• Using the data in the table calculate ∆G0, ∆H0, and ∆S0.
Make a plot of E0 versus T.
Solution:
0.051
Cell Potential (V)
0.05
0.049
0.048
0.047
0.046
0.045
0.044
0.043
0.042
290
295
300
305
310
315
T(K)
The plot is virtually a straight line with slope
∂∆E
= 3.55 × 10−4 VK −1
∂T
Assume T=298K. Then
∆G = − nℑ∆E0 = − ( 2 ) ( 96485Cmol −1 ) ( 0.0454V ) = 8761J
∂∆E
= ( 2 ) ( 96485Cmol −1 )( 3.55 × 10−4 VK −1 ) = 68.5 JK −1
∂T
∆H = ∆G + T ∆S = 8761J + ( 298K ) ( 68.5 JK −1mol −1 ) = 8761J + 20413 J = 29174 J
∆S = nℑ
•
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 :
NADH + H + + 12 O2 → NAD + + H 2O .
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:
(
)
)
−259.83kJmol -1
∆G
∆G = −nℑ∆E → ∆E = −
=−
= 2.69V
nℑ
1 96485Cmol −1
( )(
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:
C
⎛ CNAD+ ⎞
NAD +
⎜⎝ C 0 + ⎟⎠
1M
C NAD+
NAD
K=
=
=
1/ 2
1/ 2
⎛C +⎞
C NADH ⎛ C H + ⎞ PO2
C NADH ⎛ C H + ⎞ ⎛ PO2 ⎞
C NADH ⎜ CH0 ⎟ PO1/ 2
0 ⎟ 1bar
⎜
0 ⎟ ⎜
0 ⎟
0
⎜
1M
⎝ H+ ⎠ 2
C NADH ⎝ C + ⎠ ⎝ PO ⎠
⎝ CH + ⎠
2
H
( )
( ) ( )
( )
=
C NAD+
⎛C +⎞
C NADH ⎜ CH0 ⎟ PO1/ 2
⎝ H+ ⎠ 2
=
C NAD+
C NADH C H + PO1/ 2
C H0 +
2
Note if C H0 + = 1M then K =
C NAD+
C NADH C H + PO1/ 2
. We assume this definition of K in what
2
follows…
∆G ′ = − RT ln K ′ = − RT ln
⎞
⎛
C NAD+
0
0
C
=
−
RT
ln
⎟ − RT ln C H +
⎜
1/ 2
1/ 2
H+
⎜⎝ C NADH C + PO ⎟⎠
C NADH C H + PO
H
C NAD+
2
2
⎞
⎛
C NAD+
0
−7
−7
= − RT ln ⎜
⎟ − RT ln C H + = − RT ln K − RT ln10 = ∆G − RT ln10
1/ 2
⎜⎝ C NADH C + PO ⎟⎠
H
2
( )(
( )
)(
= ∆G + 7 RT ln10 = ∆G + 7 2.3 RT = −259.83kJmol −1 + 16.1 8.31JK −1mol −1 298K
= −219.96kJmol −1
−219.96kJmol -1
∆G ′
′
=−
= 2.28V
∴ ∆E = −
nℑ
1 96485Cmol −1
(
( )(
)
)
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 ∆E = ∆E +
RT
ln Q and
nℑ
RT
ln Q′
nℑ
So we have to calculate the two reaction quotients:
⎛ CNAD+ ⎞
0.004
⎜⎝ C 0 + ⎟⎠
0.004
1
NAD
=
=
= 3.23 × 104
Q=
1/ 2
1/ 2
−5
−4.8
0.035
0.05
10
0.035 1.58 × 10 0.224
C NADH ⎛ C H + ⎞ ⎛ PO2 ⎞
1
1
1
0 ⎟
0
0 ⎟ ⎜
⎜
C NADH ⎝ C + ⎠ ⎝ PO ⎠
2
H
∆E ′ = ∆E ′ −
( )
( )( )( )
( )
Q′ =
⎛ CNAD+ ⎞
⎜⎝ C 0 + ⎟⎠
NAD
C 0NADH
( )
=
( )( )( )
0.004
1
( )
C NADH
(
⎛ CH + ⎞ ⎛ PO2 ⎞
⎜⎝ C 0 + ⎟⎠ ⎜⎝ PO0 ⎟⎠
2
H
1/ 2
0.035
1
−4.8
10
10−7
0.05
1
1/ 2
=
)(
)(
)
0.004
(0.035)(1.58 × 10 )(0.224)
2
= 3.23 × 10−3 = Q × 10−7
In the ‘1M” convention:
∆E = ∆E −
(8.31JK −1mol −1 ) ( 298K ) ln 3.23 ×104
RT
ln Q = 2.69V −
(
)
nℑ
(1) ( 96485Cmol −1 )
= 2.69V − 0.27V = 2.42V
In the “10-7M” convention:
∆E ′ = ∆E ′ +
(8.31JK −1mol −1 ) ( 298K ) ln 3.23 ×10−3
RT
ln Q′ = 2.28V −
(
)
nℑ
(1) ( 96485Cmol −1 )
= 2.28V + 0.16V = 2.42V
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:
R|3Na boutsideg
gU|
+
V| + ATP → ADP + phosphate + S| +
boutsidegW
T 2 K binsideg
b
3 Na + inside
+
2K +
+
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 at 310K is ∆G 0 = −313
. kJ / mole . 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.
FG
H
Solution:
IJ
K
∆G Na
+
( in →out )
= G in − G out
⎛ ⎡ Na + ⎤
⎣
⎦ out
= n∆µ = RT ln ⎜
+
⎜ ⎡⎣ Na ⎤⎦
in
⎝
⎞
⎟ + zℑ (ψ out −ψ in )
⎟
⎠
⎛ 140 ⎞
= ( 8.31J / mol ⋅ K )( 310 K ) ln ⎜
⎟ + ( 96485C / mole )( 0 − (−0.07V ) )
⎝ 10 ⎠
⎛ 140 ⎞
= ( 8.31J / mol ⋅ K )( 310 K ) ln ⎜
⎟ + ( 96485C / mole )( 0.07V )
⎝ 10 ⎠
= 6798 J + 6754 J = 13552 J
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:
∆G K
+
(out → in )
= G in − G out
⎛ ⎡⎣ K + ⎤⎦ ⎞
in
= RT ln ⎜
⎟ + zℑ(ψ in − ψ out )
+
⎡
⎤
⎝ ⎣ K ⎦ out ⎠
⎛ 100 ⎞
+ (96485C / mole )(−0.70V ) = 1000 J / mole
= (8.31J / mol ⋅ K )(310 K )ln ⎜
⎝ 5 ⎟⎠
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, 600 J / mole ) + 2 (1000 J / mole ) = 42,800 J
(
)
(
)
g) The standard free energy change for the hydrolysis of ATP i.e.
⎛
⎞
→
ATP
at 310K is ∆G 0 = −313
. kJ / mole . If the
ADP
+
phosphate
⎜⎝
⎟
←
⎠
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:
transport
∆Gtotal = ∆Gtotal
+ ∆GATP → ADP ≤ 0
0
transport
∆Gtotal + ∆GATP → ADP = 42, 600 J + ∆GATP
→ ADP + RT ln Q ≤ 0
⎛ [P ][ADP ]⎞
transport
∆Gtotal
+ ∆GATP → ADP = 42,600 J − 31, 300 J + (8.31J / K )(310 K )ln ⎜ i
≤0
⎝ [ATP ] ⎟⎠
⎛ [P ][ADP ]⎞
transport
∆Gtotal
+ ∆GATP → ADP = 11, 500 J + (2576 J )ln ⎜ i
≤0
⎝ [ATP ] ⎟⎠
⎛ [P ][ADP ]⎞
[P ][ADP ] ≤ e−4.464 = 0.01152
ln ⎜ i
≤ −4.464 ⇒ i
⎟
[ATP ]
⎝ [ATP ] ⎠
[P ][ADP ] ≤ 0.01152 ⇒ (0.01)[ADP ] ≤ 0.01152 ⇒ [ADP ] ≤ 1.152
i
[ATP ]
Therefore,
[ATP ]
[ATP ]
[ATP ] ≤ 0.868 .
[ADP ]
7) The data below are for the binding of oxygen to the protein squid hemacyanin. The
percent saturation is the parameter
ν
N
Percent saturation
0.30
1.92
8.37
32.9
55.7
73.4
83.4
89.4
91.3
PO2 (mmHg)
1.13
7.72
31.71
100.5
136.7
203.2
327.0
566.9
736.7
× 100% = f × 100% .
Solution:
v
log
versus log PO2
1− v
log v/n-v
1.5
1
log vbar/1-vbar
0.5
0
-0.5 0
0.5
1
1.5
-1
-1.5
-2
-2.5
-3
log PO2
2
2.5
3
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 ∆H b = −75kJmol −1 .
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: qb = V ∆H b cL ( bound ) .
So we have to determine cL ( bound ) = cL ( total ) − cL ( free )
where
cL
(
) ) (
(
( )(
)(
) )
(
(
)
2
− NK bcP − K bcL total + 1 ± ⎛ NK bcP − K bcL total + 1 + 4K bcL total ⎞
⎝
⎠
free =
2K b
( )
) ( )(
)
5 × 10 M ± (25 × 10 M + 4 (5 × 10 )(5 × 10 M ))
∴ c ( free )=
2 (5 × 10 )
5 × 10 M ± (25 × 10 M + 100 × 10 )
5 × 10 M ± 5 × 10 M
=
≈
= 10
(
)
1/ 2
NK bcP − K bcL total + 1 = 4 5 × 108 10−4 M − 5 × 108 5 × 10−4 M + 1 ≈ −5 × 104 M
4
8
2
1/ 2
−4
8
8
L
4
8
2
4
1/ 2
4
109
4
109
Therefore
cL bound = cL total − cL free = 0.0005M − 0.0001M = 0.0004
(
)
∴ qb = V ∆H bcL
( ) ( )
(bound )= (0.1L)(−75kJmol )(0.0004 M )= −0.003kJ
80% of the ligand is bound.
−1
−4
M
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 ∆H b NcP = 0.1L −75kJmol −1 4 0.0001M = 0.003kJ . The protein is fully
(
saturated (100%).
)(
)( )(
)