Solutions to Haynie Ch. 4

Chapter 4

1. 1. D. The expansion expansion of a gas into vacuum vacuum is an entropic entropic phenomenon. phenomenon. There is effectively no interaction between molecules in the gaseous state, regardless of concentration. Hence, no enthalpy change. Also, ∆ H = = ∆U + + p∆ = p∆V . If q = 0, ∆ H = – p∆V + p∆ p∆V = 0. When two objects come to equilibrium, say two pieces of metal, heat is exchanged between them but there is no change in the structure of the metals – unless, of course, the coming to equilibrium results in a phase change… Also, the heat is exchanged between the objects, not the surroundings, so q = 0 for the combined system and ∆ H = = 0. 2. C. There is is an increase in entropy when two objects objects coming to equilibrium equilibrium because heat is transferred from an object at one temperature to an object at another temperature. 3. C. Work would would have to be done to compress the expanded gas, and work would have to be done to restore the objects to their original original temperatures. Thus, both processes involve a decrease in free energy. 4. D. Both changes changes are spontaneous in one direction direction only. 2. 1. A. When temperature and pressure are constant, and ∆G is negative, the process is spontaneous. A negative free energy says nothing about about reversibility. 2. D. ∆S < 0 says nothing about about the reversibility reversibility of the process. The total entropy (system plus surroundings) must increase for spontaneity, not decrease. 3. B. The work a system can do is maximal when a process it undergoes is reversible, i.e., at equilibrium. Under such conditions, the entropy change is minimal. 4. A. Dilution is spontaneous. As with the expanding gas in the previous problem, work would have have to be done to concentration the solute after dilution. 5. B. ∆S = q / T , but only if the process is reversible. When a process is reversible, measurement of T and and q enables indirect determination of ∆S . 6. B. Inefficient coupling coupling results in things things like the generation generation of heat due to friction and the like; in a word, irreversibility. irreversibility. For a reversible reversible process, i.e., a process at equilibrium, ∆G = ∆Ginput + ∆Goutput = 0. So ∆Goutput = – ∆Ginput. The efficiency of a process scales as –100%×( ∆Goutput/∆Ginput). For a spontaneous process, ∆G = ∆Ginput + ∆Goutput < 0, so ∆Goutput < – ∆Ginput, and the efficiency is less than 100%. 3. 1. C. Both U and and G are state functions, as are ∆U and and ∆G. 2. A. q + w = ∆U . This is a mathematical mathematical statement of of the First Law. 3. B. When T and and p are constant, G is the appropriate thermodynamic potential function. A potential function indicates the direction of spontaneous change. 4. D. The disorder function is S . 5. B. When a system is at equilibrium and T and p are constant, there is no tendency to change. The potential function, ∆G, must be zero. 6. B. One sums the ∆Gs of the component component reactions. If the overall change in free energy is negative, the process will be spontaneous. 7. B. ∆G measures the maximum amount of work that the system can do at constant T and and p. 4. 1. B. This comes from ∆G = 0 at equilibrium.

© 2001-2007 by D.T. D.T. Haynie. All rights reserved.

2. B. This is based on ∆µ °, which itself comes from ∆G°. 3. D. For a spontaneous change, ∆G < 0. ∆G° is equal to zero only if T = 0 or K eq = 1. 4. A. ∆ H and ∆S are not standard state functions. 5. C. Both ∆G and ∆G° are measured in units of energy mol –1. 6. A. There is no requirement that products or reactants be in the standard state. 7. B. There are two components to ∆G: a concentration independent one ( ∆G°) and one that depends on concentration. The magnitudes of the components are equal and opposite at equilibrium. 5. 1. A. The activity, or effective concentration, is defined as the product of the concentration and the activity coefficient. 2. A. ∆G = ∆G° + RT ln(a products/areactants). 3. B. nµ gives the contribution of n moles of a component of solution to the overall free energy. 6. The Sun is not absolutely necessary for life on Earth or anywhere. In general terms, all that is needed for an organism to thrive is a suitable supply of negative entropy (high free energy chemical compounds, food). Suppose the thermal energy of Earth were to be tapped as a major source of energy. In principle this is certainly possible. An effect, however, would be to decrease the temperature of the surface, requiring us to find more energy to keep warm! 7. The units depend on the chemical reaction. When the number of reactants is the same as the number of products, K eq is unitless. 8. ∆Go = – RT lnK eq o –1 –1 –1 For instance, ∆G = –8.314 J K mol × 298 K × ln(0.001) = 17.1 kJ mol K eq 0.01 0.1 1 10 100 1000 o

o

o

∆G 11.4 5.7 0 –5.7 –11.4 –17.1

o

9. ∆S = –(∆G – ∆ H ) / T –1 –1 –1 –1 = –(3.3 kcal mol + 6 kcal mol ) / 298 K = –31 cal mol K 10. K eq = exp(– ∆Go / RT ) o o = exp(–(∆ H – T ∆S ) / RT ) –1 –1 –1 –1 –1 = exp(–(150 cal mol – 298 K × 4.4 cal mol K ) / (1.9872 cal mol K × 298 K)) = 7.1 11.

B. Cellular conditions will ordinarily differ substantially from those of the standard state, and this must be taken into account when computing ∆G. Usually, ∆S data are not available. The only sensible choices are therefore B and E. The latter, however, does not take the logarithm of K .

© 2001-2007 by D.T. Haynie. All rights reserved.

12. Suppose you have the reaction A + B ⇔ C + D, and let K eq << 1. The reaction proceeds to the left – unless there is another route by which the concentration of C or D can be altered. For depletion of either C or D will result in the conversion of A and B to C and D, even if this requires getting over a large free energy barrier. Biochemical reactions are so complex because the product of one reaction is the reactant of another, possibly several others. 13. 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, [H 2O] = 55.5 M. Now, by Eqn. 4.39 and the reasoning leading to it, ∆G°' = ∆G° + nRT ln[H2O] = ∆G° + 1 mol × 8.314 J K –1 mol –1 × 298.16 K × ln(55.5 M) –1 = ∆G° + 9.96 kJ mol –1

14. ∆G°' is larger than ∆G° by about 10 kJ mol 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. Using Eqn. 4.39, ∆G°' = ∆G° + nRT ln[H2O] RT lnK eq = ∆G°' – nRT ln[H2O] ∆G° = –

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: –1

–1

–1

K eq = exp(–(–61.9 kJ mol – (–1) × 9.96 kJ mol ) / (0.008314 kJ mol × 298 8 K) = 12.7•10

The calculation is similar for pyrophosphate: –1 –1 –1 K eq = exp(–(–33.5 kJ mol + 9.96 kJ mol ) / (0.008314 kJ mol × 298 K) 3 = 13.4•10

And for glucose-1-phosphate: –1 –1 –1 K eq = exp(–(–20.9 kJ mol + 9.96 kJ mol ) / (0.008314 kJ mol × 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 (Chap. 5). 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. 15. The reaction of interest is FBP ⇔ GAP + DHAP. K eq ' = [GAP][DHAP]/[FBP] = [GAP]([DHAP]/[GAP])/([FBP]/[GAP]) = exp(– ∆G°'/ RT )

Solving for [FBP]/[GAP] gives [FBP]/[GAP] = [GAP] × 5.5 / exp(– ∆G°'/ RT ) 5

When [GAP] = 2•10 – M, we have 5

1

1

[FBP]/[GAP] = (2•10 – M × 5.5) / exp(–22.8 kJ mol – /(0.008314 kJ mol – × 5 1 310 K)) = 2•10 – M × 38,219 M – = 0.8 3

and when [GAP] = 1 •10 – M, 3

1

[FBP]/[GAP] = 1•10 – M × 38,219 M – = 40 Note that a two-orders of magnitude increase in the concentration of [GAP] can give a two-orders of magnitude increase in [FBP]/[GAP], if [DHAP] is constrained to vary proportionally with [GAP]. What sort of biochemical control mechanism can impose such a constraint? 16. In Section H we found that ∆G° for the reaction given by Eqn. 4.35 at 298 K is – 1 7.3 kJ mol – . By Eqn. 4.32, ∆G = ∆G° + RT ln([G6P]/[G1P]) –1 –1 –1 = –7.3 kJ mol + (8.314 J mol K × 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. 17. From Table 4.6, ∆V °' for this reaction is –0.19 V + 0.32 V = +0.13 V, the sum of the two relevant half-cell reduction potentials. We can calculate the effect of requiring the concentrations of oxidant and reductant to be certain values using Eqns 4.75 and 4.76. +

[lactate]/[pyruvate] = [NAD ]/[NADH] = 1


This is the trivial case. ln(1) = 0 and ∆V = ∆V ° for each half reaction; each couple is 50% reduced. As before, the potentials sum to 0.13 V, and by Eqn. 4.69 the –1 free energy change is –25.1 kJ mol . +

[lactate]/[pyruvate] = [NAD ]/[NADH] = 160 –1

–1

–1

–1

∆V lactate,pyruvate = –0.19 V – ( RT /2F )ln(160) = –0.19 V – ((8.314 J mol K × 298 –1 –1 K) / (2 × 96,494 J V mol )) × ln(160) = –0.255 V ∆V NAD+/NADH = –0.32 V – ( RT /2F )ln(1/160) = –0.32 V – ((8.314 J mol K × 298 K) / (2 × 96,494 J V –1 mol –1)) × ln(1/160) = –0.255 V

By Eqn. 4.77, ∆V = ∆V lactate,pyruvate – ∆V NAD+/NADH = –0.255 V – (–0.255 V) = 0.00 V. ∆G must be zero as well.

[lactate]/[pyruvate] = [NAD +]/[NADH] = 1000 –1

–1

∆V lactate,pyruvate = –0.19 V – ( RT /2F )ln(1000) = –0.19 V – ((8.314 J mol K × 298 K) / (2 × 96,494 J V –1 mol –1)) × ln(1000) = –0.279 V –1

∆V NAD+/NADH = –0.32 V – ( RT /2F )ln(1/1000) = –0.32 V – ((8.314 J mol –1 –1 298 K) / (2 × 96,494 J V mol )) × ln(1/1000) = –0.231 V

–1

K ×

∆V = ∆V lactate,pyruvate – ∆V NAD+/NADH = –0.279 V – (–0.231 V) = –0.048 V –1 –1 –1 ∆G = –2 × 96,494 J V mol × –0.048 V = 9.3 kJ mol +

Here, the overall reaction is not spontaneous; the NAD /NADH half-reaction has the more positive reducing potential. Lactate is now the reducing agent. +

+

Production of NAD requires ∆G < 0 or ∆V > 0. If [NAD ]/[NADH] = 1000, then ∆V NAD+/NADH = –0.231 V. Setting –1

–1

∆V lactate,pyruvate = –0.19 V – ( RT /2F )ln( x) = –0.19 V – ((8.314 J mol K × 298 K) –1 –1 / (2 × 96,494 J V mol )) × ln( x) = –0.231 V

and solving for x gives x = exp(–(–0.231 V + 0.19 V)(2 × 96,494 J V K)) = 24

–1

–1

–1

mol )/(8.314 J mol

–1

K × 298

+

The maximum value of [lactate]/[pyruvate] for production of NAD when + [NAD ]/[NADH] = 1000 is 24. 18. From Table E in Appendix C, ∆G°' for the conversion of isocitrate into α –1 + ketoglutarate is –21 kJ mol . This reaction involves NAD and NADH as shown schematically in Fig. 5.3. ∆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. From Eqn. 4.32, + –1 –1 ∆G = ∆G° + RT ln([α-KG][NADH]/([NAD ][IC])) = –21 kJ mol + 8.314 J mol –1 × 298 K × ln((0.1 mM) / 8 / 0.02 mM) = –22 kJ mol < 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. 19. If the two liquids are the same there is no change in concentration on “mixing.” There isn’t even a change in volume. If follows that there is no change in entropy. All particles interact with each other as before, so there is no change in enthalpy. It follows that there is no change in free energy. This is known as the Gibbs paradox. 20. Let the volume of one bulb be v. The entropy change is given by ∆S = Rln(2v/v) = Rln2. Assuming the gas to be relatively dilute prior to expansion (or requiring that the gas be ideal), ∆ H = 0. ∆G = ∆ H – T ∆S = – RT ln2. As required, the final state has a lower free energy than the initial one. 21. The sign of the chemical potential of species A and species B in Eqn. 4.31 is negative. These species are reactants, and we are interested in the difference in driving force between the products of the reaction and the reactants. In Eqn. 4.32 the negative sign is missing for the reactant terms because ln x –1 = –ln x, and –(– ln x) = +ln x. o

o

22. Cyt f : 3+ → 2+ V ' = 0.365 V, and Cyt c: 3+ → 2+ V ' = 0.254 V. For the reaction 2+

3+

3+

2+

Cyt c (Fe ) + Cyt f (Fe ) ⇔ Cyt c (Fe ) + Cyt f (Fe ) o

V '

= 0.365 V – 0.254 V = 0.111 V. o

o

–1

–1

∆G ' = – nF V ' = –1 × 96,494 kJ V mol

–1

× 0.111 V = –10711 kJ mol .

o

Because ∆G ' < 0, the reaction proceeds spontaneously under standard state conditions. –1

–1

–1

–1

23. ∆V = –0.32 V + (8.314 J mol K × 310 K) / (2 × 96249.4 J V mol ) × ln(1/99) = –0.38 V

24. Can any other thermodynamic function besides the Gibbs free energy predict the direction of spontaneous change under the conditions of constant temperature and pressure? Let’s see what the thermodynamics of the liquid to solid transition of


water in the vicinity of the freezing point at one atmosphere pressure can tell us. ∆U is negative above, at, and below zero, so while the magnitude of ∆U may be important, sign alone does not predict the direction of spontaneous change. Ditto for ∆ H . Same again for ∆S . The only qualitative difference between these possibilities and T ∆S is the change in sign. In contrast, ∆G has one sign where we expect spontaneous change, the opposite sign where we expect the reaction to move in the opposite direction, and is zero at the melting point. DG would be less useful to us in predicting the direction of spontaneous change if the constraints were not temperature and pressure but, say, temperature and volume or volume and pressure. Biochemists, however, almost always work under the constraints of constant temperature and pressure. 25. As we have seen, addition of an impurity will in general decrease the melting point of a pure substance. So, although the presence of cholesterol makes a membrane more rigid, it also reduces its transition temperature; the overall effect is relatively little change in melting temperature from pure lipid. Moreover, the “impurity” increases the breadth of the phase transition, meaning that it occurs over a range of temperatures. Nevertheless, a biological will be more rigid at body temperature with cholesterol present than without it. 26. Bacteria, fish, and other organisms might deal with the effect of changes in the surrounding temperature by altering the composition of cell membranes in real time. Steroids like cholesterol are one means of doing this. The rigid compound affects membrane rigidity and melting temperature. 27. An important design criterion for a liposome-based drug-delivery system is membrane rigidity. As discussed in the text and in the answer to Exercise 24 of this chapter, membrane rigidity is influenced by the length of the hydrocarbon tail and number of unsaturated bonds in a lipid and by the presence of lipo-soluble materials like steroids. One could adjust the proportions of various possible components of a membrane to make liposomes with the desired physical properties. 28. logK eq = 8.188 – (2315.5/ T ) – 0.01025 T = 8.188 – (2315.5/298) – 0.01025 × 298 = –2.637 ∆G = – RT lnK eq –1 –1 –1 = –(1.9872 cal mol K ) × (298 K) × 2.303 × logK eq = 3.596 kcal mol o

–2

∆lnK eq ≈ (∆ H / R)(T )(∆T ) ≈ 2.303∆logK eq –1

logK eq + ∆logK eq = 8.188 – 2315.5 × (T + ∆T ) – 0.01025 × (T + ∆T ) –1 –2 = 8.188 – 2315.5 × (T – T ∆T + …) – 0.01025 × (T + ∆T ) o

2

–2

∆lnK eq / ∆T ≈ ∆ H / RT ≈ 2.303 × (2315.5 T ∆T – 0.01025 ∆T ) / ∆T o

2

∆ H ≈ 2.303 × R × (2315.5 – 0.01025 T )


o

o

∆ H (25 C) –1 –1 –1 –1 2 ≈ 2.303 × 1.9872 cal mol K × (2315.5 K – 0.01025 K × (298 K) ) –1 = 6.43 kcal mol o

o

o

o

o

o

∆S (25 C) = –( ∆G (25 C) – ∆ H (25 C)) / T –1 –1 –1 –1 ≈ –(3.596 kcal mol – 6.431 kcal mol ) / 298 K = 9.51 cal mol K o

o

2

∆ H + ∆(∆ H ) = 2.303 × R × (2315.5 – 0.01025 × (T + ∆T ) ) o ∆(∆ H ) / ∆T ≈ –2.303 × R × 0.01025 × 2T o

o

o

∆C p(25 C) ≈ ∆(∆ H (25 C)) / ∆T –1 –1 –1 ≈ –2.303 × 1.9872 cal mol K × 0.01025 K × 2 × (298 K) –1 –1 = –28.0 cal mol K

29. 1. D. Neither the chemical potential of the solvent nor that of the solute equals just RT lna. o 2. C. Both chemical potentials are equal to µ + RT lna. 3. C. At equilibrium, both chemical potentials have the same value on either side of a membrane. 4. A. The chemical potential of the solute determines the osmotic pressure. This subject, however, is not dealt with until Chap. 5. 30. x

0.106 0.107 0.108

ln(1 + x) 0.101 0.102 0.103

% difference 4.95 5.00 5.04

Note: values in the table have been rounded to three significant figures.

When x is about 0.11, the difference between x and ln(1 + x) is about 5%. This provides a useful rule of thumb for making the approximation ln(1 + x) ≈ x. 31. 1. 2. 3. 4. 5.

C C B D C

32. K G = [Gln]/([Glu – ][NH4+]) = 0.00315 (M –1) The ‘M’ is given in parenthesis to emphasize that the equilibrium constant is unitless. (For further information on this, see for example Chap. 7 of Atkins, th Physical Chemistry, 6 edn.) The importance of this lies in the choice of concentration scale. For if the concentrations are in units other than mol/L, for instance mmol/L, then the argument of t he logarithm in the calculation of ∆G will be different, and ∆G will be different. But ∆G is a state function, so how could its


value depend on the choice of concentration scale? It does not. The difficulty here is that important information is not shown explicitly. And that is that each of the concentrations given in the equilibrium constant calculation must be referred to the standard state concentration, usually 1 M (as discussed in the text). –

+

K coupled = [Gln][ADP][Pi]/([ATP][Glu ][NH4 ]) = K GK ATP = 1200 –1

5

K ATP = [ADP][Pi]/[ATP] = K coupled /K G = 1200/(0.00315 M ) = 3.8 × 10 (M) ∆GATP = – RT lnK ATP = –1.9872 cal mol –1 K –1 × 310 K × ln(3.8095 × 105) = –7.9 kcal mol –1 +

–3

33. pH = –log10[H ] = –log10[10 ] = –(–3) = 3 +

+

–pH

34. pH = –log10[H ], so [H ] = 10

–6

= 10 = 1 µM 2

35. Molality is approximately the same as molarity for dilute solutions. I = 0.5Σ( zi M i) 2 2 = 0.5[ zMn × (0.35) + zCl × (0.7)] = 0.5[4 × 0.35 + 1 × 0.7] = 1.05 36. K a = [CH3COO – ][H+]/[CH3COOH] = [CH3COO – ][H+]/0.01 = 1.8 × 10 –5 –

+

–7

[CH3COO ][H ] = 1.8 × 10

– + –7 1/2 –4 [CH3COO ] = [H ] = (1.8 × 10 ) = 4.2 × 10 2

–4

2

–4

–4

–3

–3

I = 0.5[ zacetate × (4.2 × 10 ) + zhydronium × (4.2 × 10 )] = 0.5[2 × (1 × 4.2 × 10 )] –4 = 4.2 × 10

37. Assume that the temperature is 25 °C. a) 5 mM H2SO4 2

–3

2

I = 0.5[ zsulfate × (5 × 10 ) + zhydronium × (2 × 5 × 10 )] = 0.5[(4 + 2) × (5 × 10 )] = 1.5 × 10 –2

This calculation assumes complete dissociation of the acid (a reasonable approximate). Moreover, it does not include the ionization of water. The contribution, however, of hydroxyl ions and hydronium ions to the ionic strength, is negligible. 2 0.5 logγi = – Hzi I 2 0.5 –2 2 –4 logγsulfate = –0.509(2 ) I = –0.509 × 4 × (1.5 × 10 ) = –4.581 × 10 (–0.0004581) γsulfate = 10 = 0.9989 2 0.5 –2 2 –4 logγhydronium = –0.509(1 ) I = –0.509 × 1 × (1.5 × 10 ) = –1.145 × 10 (–0.0001145) = 0.9997 γsulfate = 10

b) 2 mM NaCl


2

–3

2

–3

–3

I = 0.5[ z Na × (2 × 10 ) + zCl × (2 × 10 )] = 2 × 10

logγ Na = –0.509(1 2) I 0.5 = –0.509 × 1 × (2 × 10 –3)2 = –2.036 × 10 –6 (–0.000002036) = 0.999995 γ Na = 10 γCl = 0.999995 At low ionic strength, γ ≈ 1. 38. The experimental data, when plotted, appear as shown in the figure. The first – column of numbers represents log( x/(co- x)), where x is the equivalents of OH added, and co is the equivalents of HA initially present. At the isoelectric point, – the equivalents of OH added is 0.5, and one has log(0.5/(1-0.5)) = 0. At more acidic pH, however, the logarithm is less than zero, because the numerator is smaller than the denominator. The stronger the base, the lower the pH at which 50 % dissociation occurs. So, oxyhemoglobin is a “better” acid than deoxyhemoglobin.

39. The greater the energy (heat) liberated during a reaction, the greater the affinity of the reactants for each other. As we have seen in Chapter 2, such reactions are exothermic; ∆ H < 0. Exothermic reactions are thermodynamically favourable at constant temperature and pressure because ∆G = ∆ H – T ∆S . The example of water freezing, however, shows that ∆ H alone does not predict the direction of spontaneous change. A spontaneous reaction can indeed have ∆ H > 0, which on its own would be unfavorable, but only if ∆S > 0 and T ∆S > ∆ H . As we shall see


in later chapters, the binding of one molecule to another, for example a hormone ligand to a membrane-bound receptor, can be de scribed in terms of thermodynamic quantities. One type of ligand binds a certain type of receptor with “high affinity,” another type of ligands binds the same receptor with “low affinity.” What is meant here, however, is not the enthalpy change of binding, but the free energy change, and as we have seen a negative free energy change can involve an increase in enthalpy. 40. In solution thermodynamics, ∆ H ° ≈ ∆ H except at very high concentrations of solute, when solute-solute interactions are no longer negligible. Moreover, in many cases ∆ H is approximately independent of temperature (but not for protein denaturation!). ∆G is very different: it is highly dependent on temperature and concentration. 41. The Henderson-Hasselbalch equation tells us that –

pH = pK a – log([HA]/[A ]) Rearranging to solve for the argument of the logarithm, we have –

[HA]/[A ] = 10

(pKa – pH)

4–1

= 10

= 1000 –

There are 1000 HAs for each A at pH 1, so the percentage ionized is 0.1 %. 42. Tris can be used in scanning calorimetry experiments, but it is generally advisable to avoid it if possible. Why should that be, if the heat capacity of a protein solution is always measured relative to pure buffer in a differential scanning calorimeter? If the buffer is gaining or losing protons during the experiment, the pH of solution is changing, and this could have dramatic consequences for protein stability (Chapter 2). In the ideal case, a DSC experiment will measure the effect of changing one variable only: the temperature. +

pH = pK a – log([Tris ]/[Tris]) Rearranging to solve for the argument of the logarithm, we have +

[Tris ]/[Tris] = 10

(–logKa – pH)

(–log(8.3 nM) – 8.0)

= 10

(0.0809)

= 10

= 1.2

+

There are about six Tris ions for every five Tris molecules. If the total concentration of Tris is 150 mM, then [Tris +] + [Tris] = 150 mM. We have two unknowns, but a unique determination of them can be made because we have two independent relationships involving them. Combining these gives 1.2[Tris] + [Tris] = 2.2[Tris] = 150 mM from which [Tris] = 68 mM and [Tris +] = 82 mM The increase in number of dissociated protons increases the driving force for protonation of Tris. The number of protons that become bound to Tris is the same


as the decrease in the number of unionized Tris molecules and the decrease in number of free protons, assuming that all H + from the added HCl binds to Tris (which is approximately correct). We assume complete dissociation of HCl. concentration of 10 mM,

After addition of the acid to a

+

[Tris] = 58 mM and [Tris ] = 92 mM By the Henderson-Hasselbalch equation, + –9 pH = pK a – log([Tris ]/[Tris]) = –log(8.3 × 10 ) – log(92/58) = 7.9

This is a reduction in pH by about 0.1 units. If the total concentration of Tris is reduced to 30 mM, then [Tris +] + [Tris] = 30 mM. This is a 5× dilution of the original solution. 1.2[Tris] + [Tris] = 2.2[Tris] = 30 mM +

from which [Tris] = 14 mM and [Tris ] = 16 mM After addition of 5 mM HCl, +

[Tris] = 9 mM and [Tris ] = 21 mM The pH is: + –9 pH = pK a – log([Tris ]/[Tris]) = –log(8.3 × 10 ) – log(21.36/8.63) = 7.7

This is a reduction in pH by about 0.3 units. The buffering capacity of a buffer depends on its concentration. 43. [creatineu] ≈ 40[creatines] µ = µ° + RT ln[creatine]

The transfer is from blood to urine, so ∆µ = µu – µs = µu° + RT ln[creatine]u – (µs° + RT ln[creatine]s) = (µu° – µs°) + RT (ln[creatine]u – ln[creatine] s) = (µu° – µs°) + RT ln([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 –1

–1

∆µ = RT ln([creatine]u/[creatine]s) ≈ RT ln(40) ≈ 1.9872 cal mol K × 310 K


–1

≈ 2.3 kcal mol

44. A biochemical redox reaction will be spontaneous (under standard state conditions) if ∆V °' > 0. But the reduced form of molecules in couples with negative reducing potentials are strong reducing agents, or reductants. Thus, the strongest reducing agent of the redox pairs shown in ferrodoxin. This protein, which is found in the stroma of chloroplasts, plays a key role in the light-driven transport of electrons in photosynthesis (Chapter 5).


Solutions to Haynie Ch. 4

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