Download Chem 350 – thermo problems Key 1. How many meters of stairway

Chem 350 – thermo problems
Key
1. How many meters of stairway could a 70kg man climb if all the energy available in
metabolizing an 11 g spoonful of sugar to carbon dioxide and water could be converted to
work?
The energy of oxidation of 11 g glucose = 175.2 kJ = 0.1752 kg/m2s2
(from 2870 kJ/mol *1mol/180.16g *11g)
∆x = w / (mg) = 0.1752 kg/m2s2 / 70 kg / 9.8 m s2 = 255 m.
2. A mathematical statement of the First Law of Thermodynamics is ΔU = q + w. This
holds for all processes. Assume that the only type of work done is PV work. Show
that: ΔU =+w for an adiabatic process that ΔU = 0 for an isolated system that ΔU = q for
a isochoric process that ΔH = 0 for an adiabatic process
For an adiabatic process q = 0, so ∆U = w. In addition to exchanging no heat with its
surroundings, an isolated system can do no work; hence ∆U = 0. For a process that occurs
at constant volume, w = 0, since only p-V work can be done; thus, ∆U = q. From Eqn.
2.7, ∆H = qp. If, in addition to occurring at constant pressure, the process is also
adiabatic, q = 0; giving ∆H = 0.
3. A 60 kg person drinks 0.25kg of water at 62 oC. Assume body tissues have a specific
heat capacity of 0.8 kcal/kg. By how many degrees will the hot drink raise the person’s
body temperature from the normal 37 oC. Explain how arriving at the answer involves
the First Law of Thermodyamics.
The person and the water are initially at different temperatures. When they come into
contact, heat is transferred from the water to the body until thermal equilibrium is
attained. The quantity of heat lost by the water must equal the quantity of heat gained by
the body, assuming that no heat is lost to the surroundings. The final temperature of the
body and water system, Tf, must lie between the initial temperature of the body and the
initial temperature of the water.
∆H = qp = Cp∆T = cimi∆T,
where ci is the specific heat of material i and mi is the mass of this material.
q < 0 for the water since it loses heat.
For the water, Tf – 62 oC = –q/(1.0 kcal kg–1 K–1 × 0.25 kg)
For the body tissues, Tf – 37 oC = +q/(0.8 kcal kg–1 K–1 × 60 kg)
We have two different equations and two unknowns.
Because the number of equations equals the number of unknowns, a unique solution can
be found. Rearranging the equations in terms of q, summing to eliminate q, and solving
for Tf gives 37.1 oC.
4. Explain in structural and thermodynamic terms how the unfolding of a protein is like
the melting of an organic crystal.
The unfolding of a protein is like the melting of an organic crystal in several ways. In a
crystal, individual units, e.g. individual urea molecules, are packed very close to each
other, forming a lattice. As urea molecules are uncharged, a urea crystal must be held
together by van der Waals interactions. The situation is similar in the core of a protein,
where the atoms of amino acid side chains are in close contact. Moreover, just as the
relative position of individual units is effectively constant in a crystal, so too the core of a
protein molecule is relatively rigid under native conditions. When heat is added,
vibrations increase, and when these are large enough the orderly array breaks down. If the
crystal is pure, melting occurs over a very narrow range of temperatures. Similarly, the
vibrations of amino acid side chains increase in magnitude when a protein solution is
heated. There may be larger-scale motions as well, as the atoms of a protein form a
complete, covalently-linked unit. When so much heat has been added that the van der
Waals forces are no longer great enough to prevent large-scale fluctuations, the protein
unfolds, often cooperatively. This results to a greater or lesser degree in the complete
solvation of the polypeptide chain. Exceptions to this rule do occur and these will be
discussed later.
5. For HEW (hen egg white lysozyme) calorimetric studies have shown that ΔCp,d = 1.5
kcal/molK and at pH 4.75 and ΔHd = 52 kcal/mol (at room temperature). Calculate the
enthalpy change on folding (difference between folded and unfolded state) a) at 78 oC
(the transition temperature) b) at -10 oC
∆Hd(T2) = ∆Hd(T1) + ∆Cp,d(T2 – T1).
Using the data given in the statement of the problem, we have
a) ∆Hd(78 oC) = ∆Hd(25 oC) + ∆Cp,d(78 oC – 25 oC) =52kcalmol–1
+1500calmol–1 K–1 ×53K = 130,000 cal mol–1
b) ∆Hd(–10 oC) = ∆Hd(25 oC) + ∆Cp,d(–10 oC – 25 oC) = 52 kcal mol–1 +
1500 cal mol–1 K–1 × –35 K = –500 cal mol–1
We see from part b) that the enthalpy of unfolding can be negative. This means that,
under suitable conditions heat is absorbed on protein folding. As will become clear,
however, the entropy of folding is so favorable under such conditions that the protein
does fold.
6. When glucose is combusted 6.73 kcal are given off per mole of glucose oxidized at 25
oC. a) What is ΔU at this temperature? b) What is ΔH at this temperature?
c) Suppose glucose is fed to a culture of bacteria and 400 kcal/mol of glucose is
catabolized. Is there a discrepancy between your answer for b? If so explain.
H=∆U+∆(nRT)
∆U=–673kcalmol–1 ∆n=0, so ∆H= –673kcal mol–1.
The bacteria give off less heat in consuming glucose than is measured in a bomb
calorimeter because they do not completely oxidize the carbohydrate to carbon dioxide
and water; some of the combustion energy is used for growth and therefore not given off
to the environment as heat.
7. A protein called lactalbumin is a close homolog of hen egg white lysozyme. Unlike
lysozyme, lactalbumin binds Ca2+ with high affinity. The measured enthalpy of binding,
however, is much smaller in magnitude than the enthalpy of hydration. Explain.
Analysis of the crystallographic structure of baboon α-lactalbumin has revealed that
several water molecules remain bound to the calcium ion in the binding site. That is, the
binding process removes only some of the water molecules in the solvation shell. The
enthalpy of solvation of a particular water molecule depends on how many molecules are
already bound to the ion. The enthalpy change is greatest for the first water molecule and
progressively smaller for subsequent ones. The enthalpy difference between the bound
and unbound states of calcium in α-lactalbumin is therefore only a few kilocalories per
mole.
Second Law
8. Suppose 45 J is transferred from a heat source at 375 K to a heat sink at 25 oC.
Calculate the maximum work that can be done.
Efficiency of a heat engine = wmax/qtransferred
= (1 – Tcold/Thot)
= 1 – 298K / 375K = 0.205.
wmax = qtransferred × efficiency = 45 J × 0.205 = 9.2 J.
9. One calorie is produced for every 4.1840 J of work done. If 1 cal of heat is available
can 4.184 J of wok be accomplished with it? Explain.
No. The transfer of x J of heat cannot produce x J of work. Heat transfer is always
attended by an increase in entropy, which corresponds to the fraction of heat transferred
that is unavailable to do work.
10. (3.8) For a protein, suppose that DHd =10 kcal/mol at room temperature. The
melting temperature is 68 C and the constant pressure heat capacity ΔCP = 1650cal/molK.
Calculate the following quantities and explain their significance:
a) ΔSd at Tm
b) ΔSd at 37 C
c) ΔSd at 15 C
d) At what temperature will ΔSd =0?
∆Hd(68 oC) = 10,000 cal mol–1 + 1,650 cal mol–1 oC–1 × (68 oC – 25 oC)
= 80,950 cal mol–1.
∆Sd(Tm) = ∆Hd(Tm)/Tm
= 80,950 cal mol–1 / (68 K + 273 K) = 237 cal mol–1 K–1.
This is the increase in entropy on protein unfolding.
∆Sd(37 oC) = ∆Sd (Tm) + 1,650 cal mol–1 K–1× ln(310 K / 341 K)
= 237.4 cal mol–1 K–1– 157.3 cal mol–1 K–1
= 80.1 cal mol–1 K–1
The change in entropy is strongly dependent on temperature.
∆Sd (15 oC) = ∆Sd (Tm) + 1,650 cal mol–1 K–1× ln(288 K / 341 K)
= 237.4 cal mol–1 K–1– 278.2 cal mol–1 K–1
= –41.3 cal mol–1 K–1
o
At 15 C, the entropy change for this protein is less than zero. In other words, there is an
decrease in entropy on protein unfolding! This would probably come about from an
increase in the order of water molecules solvating hydrophobic amino acid side chains.
0 = ∆Sd (Tm) + 1,650 cal mol–1 K–1 × ln(T / 341 K) T
= 341 K × exp(–237.4 cal mol–1 K–1/ 1,650 cal mol–1 K–1)
= 295 K.
At around 295 K, the various contributions to the entropy of unfolding cancel each other
out; there is no difference in entropy between the folded state and the unfolded state.
11. Is it possible for heat to be taken into a system and converted into work with no other
change in the system or surroundings? Explain
If the heat transfer is reversible, ∆S = q/T (∆S > q/T when the transfer is irreversible).
The quantity of heat transferred, q, is positive for the system but negative for the
surroundings.
So the total change in entropy is ∆Ssystem + ∆Ssurroundings = q/T + (−q)/T = 0;
there is no entropy change for a reversible transfer heat at constant temperature.
This tells us that because the inside of a living organism is effectively uniform in
temperature, exchanges of heat within the organism cancel out and result in no change in
entropy. The situation is different at the system-surroundings border, which in the case of
humans is the skin. Here heat is readily lost to the surroundings. To maintain body heat,
we must eat
12.What are the units of Keq? Explain
The units depend on the chemical reaction. When the number of reactants is the same as
the number of products, Keq is unitless.
13. Calculate DGo for Keq values of 0.001,0.01, 0.1, 1, 10, 100 and 1000. Which
reactions are spontaneous at STP?
∆Go = –RTlnKeq.
For instance, ∆Go = –8.314 J K–1 mol–1 × 298 K × ln(0.001) = 17.1 kJ mol–1.
Keq
∆Go
0.01
11.4
0.1
5.7
1
0
10
-5.7
100
-11.4
1000
-22.8
14. (4.10) When a photon in the visible range is absorbed in the retina by rhodopsin, the
photoreceptor in rod cells, 11-cis-retinal is converted to the all trans isomer. Light energy
is transformed into molecular motion. The efficiency of photos to initiate the reaction is
about 20% at 500 nm (57kcal/mol). About 50% of the absorbed energy is available for
the next signaling step. This process takes about 10 ms. In the absence of light,
spontaneous isomerization of 11-cis-retinal is very slow, on the order of 0.001 1/yr.
Experimental studies have shown that the equilibrium energetics of retinal isomerization
are DSo =4.4 cal/molK and DHo = 150 cal/mol. Calculate the standard state free energy
and the equilibrium constant.
Keq
= exp(–∆Go / RT)
= exp(–(∆Ho – T∆So ) / RT)
= exp(–(150 cal mol–1 – 298 K × 4.4 cal mol–1 K–1) / (1.9872 cal mol–1 K–1 × 298 K))
= 7.1.
15. Calculate DG when the concentrations of glucose-1-phosphate and glucose-6phosphate are maintained at 0.01mM and 1mM, respectively. Comment on the meaning
of the sign of DG and what that means to the reaction you have written.
∆G° for the conversion for the conversion at 298 K is –7.3 kJ mol–1 . This value can be
obtained from the net
∆G°' = ∆G° + RTln([G6P]/[G1P])
= –7.3 kJ mol–1 + (8.314 J mol–1 K–1•298 K•ln(1 mM/0.01 mM)
= 4.1 kJ.
The direction of the reaction has been reversed by adjusting the concentrations of
reactants and products. Control of the concentration of a metabolite is an important
means of regulating the flux of matter through biochemical reaction pathways.
16. Show that for a reaction at room temperature that yields 1 mol of water
ΔGo’ = ΔGo + 9.96 kJ/mol (remember [H2O] in an aqueous solution is 55.5 M)
We require that the reaction occur in dilute aqueous solution and yield n H2O molecules:
A + B ⇔ C + D + nH2O
Because the solution is dilute, [H2O] = 55.5 M
∆G°' = ∆G° + nRTln[H2O]
= ∆G° + 1 mol•8.314 J K–1 mol–1 • 298 K•ln (55.5 M)
= 9.96 kJ mol–1.
17. The equilibrium constant for
Glu- + NH4+ <=> Gln + H2O
is 0.00315 1/M at pH 7 and 310 K; the reaction lies far to the left.
The synthesis of gln from glu is made energetically favorable by coupling it to hydrolysis
of the terminal phophodiester bond of ATP. The equilibrium constant for the coupled
reaction, which is known from experiments with glutamine synthase is 1200. Calculate
the phosphate bond energy in ATP at pH7 and 310 K.
KG = [Gln]/([Glu– ][NH4 + ])
= 0.00315 M–1.
Kcoupled = [Gln][ADP][Pi]/([ATP][Glu–][NH4+])
Combining the two reactions produces an equilibrium that is the product of the two
individual reactions
= KGKATP
= 1200.
KATP = [ADP][Pi]/[ATP] = Kcoupled/KG
= 1200/(0.003 15 M–1)
= 3.8 × 105 M.
∆GATP = –RTlnKATP
= –1.9872 cal mol–1 K–1 × 310 K × ln(3.8095 × 105 )
= –7.9 kcal mol–1.
18. The concentration of creatine in urine is 40 fold greater than in serum. Calculate the
free energy change per molecule required for transfer of creatine from blood to urine at
37 oC.
[creatineu] ≈ 40[creatines]
G = G° + RTln[creatine]
The transfer is from blood to urine,
so ∆G = Gu – Gs
= Gu° + RTln[creatine]u – (Gs° + RTln[creatine]s)
= (Gu° – Gs°) + RT(ln[creatine]u – ln[creatine]s)
= (G u° – Gs°) + RTln([creatine]u/[creatine]s).
We assume that the standard state free energy of creatine does not depend on whether it is
in urine or serum. This is not a bad approximation because creatine is soluble and the
concentration of water is high in urine and serum.
So we are left with ∆G = RTln([creatine]u/[creatine]s)
≈ RTln(40)
≈ 1.9872 cal mol–1 K–1•310 K
≈ 2.3 kcal mol–1
19. Calculate Keq for the hydrolysis of the following compounds at neutral pH and 25C:
phophoenolpyruvate ΔGo’ = -61.9 kJ/mol
pyrophosphate ΔGo’ = -33.5 kJ/mol
glucose-1-phosphate ΔGo’ = -20.9 kJ/mol
(you can find the needed reaction by looking up glycolysis)
∆G°' is larger than ∆G° by about 10 kJ mol–1 when one mole of water molecules is
produced by the reaction. This must be taken into account because our known quantities
are primes and the unknowns are not.
∆G°' = ∆G° + nRTln[H2O]
∆G° = –RTlnKeq = ∆G°' – nRTln[H2O].
In hydrolysis water molecules are consumed, so n, the number of water molecules
produced, is negative.
The hydrolysis of phosphoenolpyruvate yields pyruvate and phosphate, and one molecule
of water is consumed.
Using the result of the previous problem and plugging in known values gives:
Keq = exp(–(–61.9 kJ mol–1 – (–1)•9.96 kJ mol–1) / (0.008 314 kJ mol–1•298 K)
= 12.7•108 .
The calculation is similar for pyrophosphate:
Keq = exp(–(–33.5 kJ mol–1 + 9.96 kJ mol–1) / (0.008 314 kJ mol–1•298 K)
= 13.4•103 .
And for glucose-1-phosphate:
Keq = exp(–(–20.9 kJ mol–1 + 9.96 kJ mol–1) / (0.008 314 kJ mol–1•298 K)
= 82.7.
Note that the driving force for hydrolysis of phosphoenolpyruvate is much greater than
that of glucose 1-phosphate. Phosphoenolpyruvate is a product of Reaction 9 of
glycolysis, the metabolism of glucose. Glucose 1-phosphate is a direct product of
glycogen breakdown. Conversion to glucose 6-phosphate enables the sugar to enter the
glycolytic pathway. Pyrophosphate is a product of DNA synthesis (and other biochemical
reactions), as only one phosphate group of a nucleotide is incorporated into the sugar–
phosphate backbone of the nucleic acid.
20. The citric acid cycle is the common mode of oxidative degradation in eukaryotes and
prokaryotes. Two components of the citric acid cycle are a-ketoglutarate and isocitrate.
If
[NADox]/[NADred] = 8
[a-ketoglutarate] = 0.1mM
[isocitrate] = 0.02mM
Assume 25 C and pH 7.0.
a) Calculate ΔG.
b) Is this reaction a likely site for metabolic control? Explain
∆G°' for the conversion of isocitrate into α-ketoglutarate is –21 kJ mol–1.
This reaction involves NAD+ and NADH.
∆G°' = ∆G° because water is not involved in the reaction; there is no change in pH if it is
assumed that the carbon dioxide produced by the reaction does form carbonic acid.
∆G = ∆ G ° + R T ln([α-KG][NADH]/([NAD+ ][IC]))
= –21 kJ mol–1 + 8.314 J mol–1•298 K•ln((0.1 mM) / 8 / 0.02 mM)
= –22 kJ mol–1 < 0.
This reaction is a site for metabolic control because ∆G is negative. Moreover isocitrate
dehydrogenase, the enzyme that catalyses this reaction, is strongly inhibited in vitro by
NADH, a product of the reaction. In contrast, most of the other reactions of the citric acid
cycle have ∆G ≈ 0.