Unformatted text preview: 20.110/7.10 and 5.60
Fall 2013
Exam #2 Solution Key
•
•
•
•
•
•
•
• For each problem, please put a square around your final answer.
Make sure that your answers show the units. Points will be taken off for incorrect or missing units.
Show your work; we give as much partial credit as possible even if you do not fully answer the question.
You may use 1 page of notes (double sided 8 ½ x 11) and a calculator that cannot communicate or take photos.
Questions will not be answered during the exam.
The exam is worth 100 points.
The last page contains constants/math/tables/etc.
This exam has 9 pages. 1. A “phase change material” is a substance with a high heat of fusion that is capable of releasing and storing
large amounts of energy upon melting and solidifying at a certain temperature. As such, they have diverse
applications, from consumer goods (e.g., water coolers) to industrial and specialized niches (e.g., spacecraft
thermal systems and solar power plants). Using a phase change material can be much more efficient than
directly raising temperature of a substance (for example, liquid water) to absorb heat. Lauric acid (C12H24O2) can be a good phase change material. The following are properties of this material:
Melting point:
Density of solid:
Density of liquid: 317.4 K (the pressure at the melting point is 100kPa)
1007 kg/m3
862 kg/m3 At 318.15 K, the pressure of liquidsolid coexistent lauric acid is 3289 kPa. a. (10
1. aLpts) Determine the heat of fusion ΔfusH of 1 kg of lauric acid.
Use the Clapeyron Equation :
dP = dT Dfus H
T Dfus V Integrate : P2  P2 = Dfus H
Dfus V ln T2
T1 , Note : Dfus V = 1
rliquid  1
rsolid Solve and substitute :
Dfus H = HP2  P2 L Dfus V
T2 H3289 kPa  100 kPaL J
= ln J N ln J T1 1
862 kg m3 318.15 K
317.4 K  1
1007 kg m3 N
= 225.7 kJ kg1 INote : 1 J = 1 Pa m3 M N 1. bL P,
Assume Dfus H is independent of T. At constant 1 DG2  DG1 = DH 1  1 Ø DGfus H320 K, 100 kPaL  DGfus H317.4 K, 100 kPaL = DHfus 1  1 dP 1. aL Dfus H = T Dfus V dT Use the Clapeyron Equation :
Name___________________________________________ TA_________________________________ Integrate : dP = dT Dfus H P2  P2 = Dfus H T Dfus V Dfus V ln T2
T1 , Note : Dfus V = 1
rliquid  1
rsolid b. (10 pts) An uninterruptible power supply is an electrical apparatus that provides emergency power
the main power fails. It often has to perform in an outdoor environment, where the temperature
Integrate : when
Solve and substitute :
can vary over a large range. Phase change materials are often used to maintain a stable working
1
1
kPa
100 kPaL
device and
N the internal pressure is 100kPa,
temperature.
Given that there isH3289
1kg
of lauric
acidJ 862
inkgthe
m3
1007 kg m3
Dfus H
T2 HP2  P2 L Dfus V
1
1
1
= 225.7
kJ kg
H
=
INoteis: 320
1 J =K.1 Pa m3 M
=
D
determine
ΔG
for
the
melting
of
lauric
acid
if
the
temperature
inside
the
power
supply
fus
P2  P2 =
, NoteT: Dfus V =
ln
Dfus V ln J 2 N T1 318.15 K rliquid T1 rsolidln J 317.14 K N 1. bL Solve and substitute :
Assume Dfus H is independent of T. At constant P, HP2  P2 L Dfus V Dfus H =DG 2 T2 DGT12 H3289 kPa  100 kPaL J =
1 1 ln J T N= DH
Ø
T2 T1
T1 1 1 862 kg m3 DGfus H320 K,
100KkPaL
318.15 ln J 320317.14
K K N  1
1007 kg m3 N kJ kg1 INote :1 1 J = 1 1Pa m3 M
DGfus H317.4=K,225.7
100 kPaL
= DHfus
317.4 K
320 K 317.4 K 1. bL
DGfus = 0 at of
coexistance,
Assume Dfus HSince
is independent
T. At constant P, DG2
T2  DG1 DGfus H320 K, 100 kPaL = DH T1 0 1 1 1
DGfus H320 K,=100
kPaL
DGfus H317.4
1
1
H225.7
kJ ê kgL
 K, 100 kPaL
K Ø
= DHfus
320
317.4 K
320 K 317.4 K
T2 T1
320 K
317.4 K
320 K 317.4 K
1 DGfus H320 K, 100 kPaL = 1.849 kJ kg1 Since DGfus = 0 at coexistance,
1. cL DGfus H320 K, 100 kPaL  0 = H225.7 kJ ê kgL 1  1 1
Since DGfus H320
K, 100
melting,
320 K
317.4
K kPaL = 1.849 kJ kg320<K0, the317.4
K and absorption of heat, will be spontaneous. DGfus H320 K, 100 kPaL = 1.849 kJ kg1 c. (5 pts) In part b, will the absorption of heat by the lauric acid be spontaneous?
1. cL
Since DGfus H320 K, 100 kPaL = 1.849 kJ kg1 < 0, the melting, and absorption of heat, will be spontaneous. 2 2. Collagen is the most abundant protein in the human body. It is a fibrous protein that serves to strengthen and support tissues. Suppose a collagen fiber can be stretched reversibly with a force constant of k = 10 N/m and the force F = k×L. When a collagen fiber is contracted reversibly at constant temperature T = 310 K from L = 0.2 to 0.1 cm, it absorbs heat qrev = 0.05 J. a. (10 pts) Calculate the change in the Helmholtz energy ΔA. 2. aL Determine what the Helmholtz energy is at constant T :
DA = wrev HdA = SdT  pdV, so at constant T, dA = pdV, and DA = wrev L Determine work :
= aL
wrev = ‡ F „ x 2.
‡ kx „ x = 1
2 kx2 Determine what the Helmholtz energy is at constant T :
Substitute and solve :
0.1 cm
1DA = w HdA = SdT  pdV, so at constant T, dA = pdV, and DA = w L
5 rev
1
2
2
rev cm
F „[ kx2 †x=0.1
J
wrev = ‡
x=0.2 cm = H0.5L µ I10 N m MAH0.001 mL  H0.002 mL E = 1.5 ¥ 10
0.2 cm
2 2. bL Determine work : 1
wrev = ‡ F „ x = ‡ kx „ x = kx2
2
Application of the First Law : andµsolve
DU = q + w = Substitute
0.05 J  I1.5
105 M:J = 0.049985 J
0.1 cm
1
5
cm
1
2
2
F „ x kx2 †x=0.1
J
wrev = ‡
x=0.2 cm = H0.5L µ I10 N m MAH0.001 mL  H0.002 mL E = 1.5 ¥ 10
0.2 cm
2
b. (10 pts) Calculate the change in internal energy ΔU. 2. bL
Application of the First Law :
DU = q + w = 0.05 J  I1.5 µ 105 M J = 0.049985 J 3 Name___________________________________________ TA_________________________________ 3. A common model of protein folding is a two state model, where the native and unfolded states exist in equilibrium. For example, the following reaction captures the folding of SBP (soybean seed coat peroxidase). SBPunfolded SBPfolded Below are the data collected from an experiment measuring the ratio of the protein in its folded to unfolded states [SBPfolded]/[SBPunfolded] at different temperatures. Temperature (oC) 47 57 67 77 87 Fraction folded 1.5x104 1.7x103 2.2x102 3.1x101 5.0x100 Use the following graph paper to solve parts a and c. You need to show the plots for full credit. Don’t over‐worry about super‐straight lines or perfect points! 4 a. (15 pts) Calculate ΔH and ΔS for the folding reaction (unfolded to folded). 3. aL lnIKeq M =  slope = DHfold DSfold + R RT DAln IKeq ME H9.62  1.61L = 1 = 23 017 K H0.003125  0.002777L DI M
T We use slope to calculate DHfold :
slopeHmL = 23 017 K =  DHfold
R ö DHfold = I8.314 J mol1 K 1 M µ H23 017 KL = 191.3 kJ mol1 Using any point and knowing the slope m, We can solve for the intercept and consequently DS Using 1
T , lnIKeq M = I0.003125 K 1 , 9.62M HNote : any point could be used, giving slightly different solutions due to rounding differencesL lnIKeq M =  DSfold DHfold + DSfold
R RT ö 9.62 = H23 017 KL I0.003125 K 1 M + DSfold
R = 62.31 ö DSfold = I8.314 J mol1 K 1 M µ H62.31L = 518.0 J mol1 K1 R 3. bL ln K2 = K1 DH 1 R T2  1
T1 Using data at 47 and 87 °CH320 K and 360 KL
R ln J
DH = J 1
T2  K2
K1
1
T1 I8.314 J mol1 K 1 M ln J N
= I N Again, lnIKeq M =  DHfold 360 K DSfold
R RT So, DSfold = I8.314 J mol + 1 1 1  1
320 K 2.73µ102
8.22µ105 N
= 191 795 J mol1 º 191.8 kJ mol1 M ö DSfold = RlnIKeq M + 5 K M lnI8.22 µ 10 M + T 191 795 J mol1
320 K 5 3. cL DHfold = 486.1 J mol1 K1 DHfold ö DHfold = I8.314 J mol1 K 1 M µ H23 017 KL = 191.3 kJ mol1
R
Name___________________________________________ TA_________________________________ slopeHmL = 23 017 K =  Using any point and knowing the slope m, We can solve for the intercept and consequently DS 1 contain disulfide bonds that are sulfur linkages between cysteine amino acids. When a protein
b.
(10 pts)Using
Proteins
, lnIKeq M =
unfolds, these linkagesTremain. The SBP protein has two cysteine amino acids that form such a bond. If one of these
cysteines is mutated,
then the
data
is measured.
Calculate
the new
ΔHdifferent
and ΔS solutions
for the folding
of this protein.
, 9.62M HNote
: any
point could be
used, giving
slightly
due to rounding
differencesL
I0.003125
K 1following Temperature (K) [SBP
/[SBP mut, folded
mut, unfolded
DSfold
5 DSfold
8.22x10
K 1 M +
+ 47 ö 9.62 = H23 017 KL I0.003125
lnIKeq M = R
RT
57
9.20x104 R
67
1.18x104
77
1.70x103
DSfold
1 1
2
= 62.31 ö DS
= 518.0
J mol1 K1
87fold = I8.314 J mol K M µ H62.31L
2.73x10
R DHfold 3. bL ln K2 = K1 DH 1 R T2  1
T1 Using data at 47 and 87 °CH320 K and 360 KL
R ln J
DH = J 1
T2  K2
K1
1
T1 I8.314 J mol1 K 1 M ln J N
= I N Again, lnIKeq M =  DHfold + 360 K DSfold
R RT So, DSfold = I8.314 J mol 1 1 1  1
320 K 2.73µ102
8.22µ105 N
= 191 795 J mol1 º 191.8 kJ mol1 M ö DSfold = RlnIKeq M + 5 K M lnI8.22 µ 10 M + DHfold
T 191 795 J mol1
320 K = 486.1 J mol1 K1 3. cL
DDS = DSmutant  DSwild type = 486.1 J mol1 K 1 + 518.0 J mol1 K 1 = 31.94 J mol1 K 1
DDH = DHmutant  DHwild type = 191.8 kJ mol1 + 191.3 kJ mol1 = 0.4 kJ mol1 º 0 kJ mol1
INote : a single weak hydrogen bond is worth about 8 kJ mol1 ,
so you can see that 0.4 kJ mol1 is essentially nothing. Also, our starting data has only 2 sig figsM 6 J T2  T1 I N Again, lnIKeq M =  DHfold DSfold
R RT So, DSfold = I8.314 J mol c. + 360 K 1 1  320 K M ö DSfold = RlnIKeq M + 5 K M lnI8.22 µ 10 M + DHfold
T 191 795 J mol1
320 K = 486.1 J mol1 K1 (5 pts) Use parts a and b to calculate ΔΔS and ΔΔH that occurs due to the mutation. In terms of microstates
rationalize your answer.
3. cL
DDS = DSmutant  DSwild type = 486.1 J mol1 K 1 + 518.0 J mol1 K 1 = 31.94 J mol1 K 1
DDH = DHmutant  DHwild type = 191.8 kJ mol1 + 191.3 kJ mol1 = 0.4 kJ mol1 º 0 kJ mol1
INote : a single weak hydrogen bond is worth about 8 kJ mol1 ,
so you can see that 0.4 kJ mol1 is essentially nothing. Also, our starting data has only 2 sig figsM First, let ' s rationalize DDH. We found that DDH º 0,
which suggests that DHmutant is about the same as DHwild type . For both the mutant and wild type protein, DH represents Hfolded Hunfolded . The relative enthalpies of these configurations are dictated by the strengths and numbers of non covalent interactions among the amino acid side chains and between the side chains and solvent molecules. Unfolded,
both the mutant and wild type proteins have roughly the same potential to interact with solvent—
Hmutant, unfolded º H wild type, unfolded . Assuming that both the mutant and wild type proteins
assume the same conformation Hsame side chains interactL when folded Hjust missing the S  S bondL,
then the non  covalent interactions between side chains should be the same : Hmutant, folded º H wild type, folded . Thus,
DHmutant is about the same as DHwild type , and as observed, DDH º 0.
Rationalizing DDS requires discussion of microstates. Generally,
the larger the number of structural configurations available to the protein Hthe number of microstatesL,
the greater the absolute entropy. This follows directly from the statistical definitions of entropy. Since we observe DDS =
31.94 J mol1 K 1 > 0, the magnitude of the entropy change of the mutant during folding is
less than that of the wild type protein : †DSmutant § < °DSwild type •. In the unfolded configurations,
both the mutant and wild type proteins have many, many possible configurations HmicrostatesL available to "explore,"
largely dictated by configurations involving solvent and neighboring protein. Thus, the absolute entropies
of both unfolded proteins are comparable : Smutant, unfolded º Smwild type, unfolded .In the folded configurations,
restriction in the number of microstates results from self
interactions : entropy differences relate to internal degrees of freedom. The wild type protein has the disulfide present,
and this covalent bond "locks" the orientations, relative angles, and distance separating domains of the protein. The
mutant does not have this additional restriction in the folded configuration since the disulfide bond is not present. Thus,
we expect the number of folded "microstates" Hdegrees of freedomL of the wild type protein to be
fewer than for the mutant. That is, Smutant, folded > Swild type, folded . Since DS = Sfolded  Sunfolded ,
we thus expect that DSmutant to be less negative than DSwild type I†DSmutant § < °DSwild type •M. This explains why DDS > 0,
as observed. 7 Name___________________________________________ TA_________________________________ 4. An RNA oligonucleotide has the sequence A4C7U4. It can form a hairpin loop held together by a maximum of four A‐U base pairs. The C nucleotides cannot form a base pair with either A or U. The A nucleotides cannot bind to themselves. The U nucleotides also cannot bind to themselves. For the following questions, assume that no “bulges” can form. That is, every structure has a single contiguous set of A‐U bonds. Note that each A can bind to different U’s to form various structures (not just the U that it is partnered with in the image). The following represents the structure of the ground state: Each A‐U line is a “base pair” and the “‐C” are unpaired (and cannot pair with A or U). The energy of a single A‐U base pair is EAU. The answers for 4c‐e will be functions of EAU and T (temperature). a. (5 pts) Draw the energy levels and assign an energy to each level. 4. aL 3 EAU Energy Energy 4 EAU
2 EAU 3 EAU 1 EAU 2 EAU 0 1 EAU 0 b. (5 pts) Calculate the degeneracy of each energy level. 4. bL
WH0L = 1
WH1 EAU L = 2
WH2 EAU L = 2
WH3 EAU L = 2
WH4 EAU L = 1 4 EAU 4. cL 8 Ei Q = ‚ Wi e k B T E AU
=1
4. bL WH0L EAU L = 2
1 EWH1
AU
L=2 WH2 E 4. bL WH0L0 = 1 AU
WH3 E L = 2
WH1 EAU L = 2AU
WH4 EAU L = 1
WH2 EAU L = 2
WH3 EAU L = 2
c. (5 pts) Write the canonical partition function for an individual RNA molecule. WH4 EAU L = 1
4. cL WH0L = 1
WH1 EAU L = 2
WH2 EAU L = 2
WH3 EAU L = 2
WH4 EAU L = 1 Ei Q = ‚ Wi e k B T
i 4. cL EAU Ei Q=1+2e 2 EAU +2e kB T Q = ‚ Wi e k B T
i
EAU 4. cL 4. 2dLe
Q=1+ +2e +2e
3 EAU +2e kB T 4 EAU +e kB T kB T 2 EAU kB T 3 EAU kB T 4 EAU +e kB T kB T d. (5pts) Calculate the probability of the ground state. E
i The Ground state refers to the lowest energy state; E = 0 : Q = ‚ Wi e k B T
i
EAU 4. dL2 E Q = 1 + 2 e kB T + 2 e Ei Wi e AU T
+ 2 ePikB=
+e kB T 1 T
4kBEAU 3 EAU kB T = Q to the
Q lowest energy state; E = 0 :
The Ground state refers
Ei 4. dL =
P4.i eL Wi e 1 kB T = Q energyQstate; E = 0 :
The Ground state refers to the lowest Wi e k B T 1 The average energy is defined as : XE\ =
‚ pi Ei = ‚
=
Pi =
i
i
Q
eL Q the average energy of an RNA molecule. e. (5 pts) 4.
Calculate Ei Wi e B Ei Q The average energy is defined as : XE\ = ‚ pi Ei = ‚ Here ‚ pi Ei =
i 1
Q ‚ pi Ei =
i = 41EAU e Q EAU
kB T i i + 4E 0 + 2 EAU µ 2 e EAU 4 EAU e 4 EAU e kB T
= + 4 EAU e = 3 E e
+ 4 EAU AU kB T kB T AU + 6 EAU e kB T kB T i B T kB T i µ2 e
+ 3 EEAU
Q 3 EAU + 6 EAU e kB2
T EAU + 2 EAU µ 2 e + 3 EQAU µ 2 e kB T + 4 EAU e 4 EAU + 4 EAU µ e + 6 4EEAU e
AU kB T kB T 4 EAU + 4 EAU e kB T kB T 9 + 4E 3 EAU + 3 EAU µ 2 e 3 EAU
kB T kB T Q Q kB T 4 EAU 3 EAU + 4 EAU e 3 EAU Ei 2 EAU k T
2 E B EAU e 2 EAU + 2 EAU µ 2 e i 2 EAU EAUk T
B
AU
kB T EAU 0 + 2 EAU µ 2 e kB T Wi e k B T
Q W e 2 EAU i EAU The average
i=
Ei = is0 defined
+ 2 EAUasµ:2XE\
e k T= ‚
+ 2piEEAU
µ‚
2e k
‚ pi energy Here Ei Ei 1 4. eL kB T Q Here i Ei kB T + 4 EA Name___________________________________________ TA_________________________________ Parameters and Conversions: Gas Constant: R = 8.314 J / mol‐K = 0.082 L atm/mol‐K = 1.986 cal/mol‐K = 8.314 L kPa/mol‐K Temperature: [oC] = [K] ‐ 273 Avogadro’s Number: 6.02 x 1023 Pressure: 1 Pa = 10‐5 bar = 1 N/m2 = 9.87x10‐6 atm Energy: 1 J = 1 N‐m = 1 kg(m/s)2 Math !"! = 1 !
! 2 10 ...
View
Full Document
 Fall '08
 Prof.KeithA.Nelson
 Trigraph, WI, EAU, T EAU

Click to edit the document details