Chem340 Physical Chemistry for Biochemists Dr. Yoshitaka Ishii Homework 10 Please solve the following questions. This is a good part of the final Exam. P9.13-14, P9.18, P9.21, P9.24, P9.28-9, P9.32, P9.40, P9.44 P9.48, P9.50 P10.1-10.2, P10.4-10.7, P10.9 P9.13) In the Debye-Hückel theory, the counter charge in a spherical shell of radius r and thickness dr around the central ion of charge +q is given by –q2re–r dr. Calculate the most probable value of r , rmp, from this expression. Evaluate rmp for a 0.050 m solution of NaCl at 298 K. d   q 2 re  r  dr   q 2 e 1   rmp  0; rmp     rmp  q 3 rmp e  rmp 0 1   I/I mol kg-11  solvent 8 0.05000.997 1 8 1 0.050 mol kg  9.211  m 1 1 1 7.33  10 m 8 10 7 m = 9.211 10 m 2.32 10 m    78.54 78.54  εr 10 κ1 29.211 .9110  108 rmp r r1mp 14.30 10 m . 36 10 m =43.0 1 . 36nm nm 8 7 κ P9.14) Calculate the Debye–Hückel screening length 1/ at 298 K in a 0.00100 m solution of NaCl. For water, the screening length at 298 K in m-1 can be calculated as: κ  2.91  1010   I/ mol kg -1  solvent  9.211  10 8 εr 0.00100.997 78.54 m 1  1.04  10 8 m 1 1  9.65  10 9 m  9.65 nm κ P9.18) Calculate I, ±, and a± for a 0.0250 m solution of K2SO4 at 298 K. Assume complete dissociation. How confident are you that your calculated results will agree with experimental results? K 2SO 4  v  2, z  1, v  1, z  2 0.0250  2  4  0.0750 mol kg 1 2 ln    1.173  2  0.0750  0.6425 I    0.523 m  a       m  m3   0.050  0.0250  6.25  105 2 m  0.03969 mol kg 1 a  0.03969  0.5260  0.0209 Not very confident. Figures 10.5 and 10.6 show significant deviations from predicted behavior for I > 0.01 mol kg–1. P9.21) Dichloroacetic acid has a dissociation constant of Ka = 3.32  10–2. Calculate the degree of dissociation for a 0.125 m solution of this acid (a) using the Debye–Hückel limiting law and (b) assuming that the mean ionic activity coefficient is one. a) We first consider the case when  is given by the Debye–Hückel limiting law. The ionic strength is given by m  2  m  0.04992 mol kg 1 2 ln    1.173  1  0.04992  0.2621 I    0.7694 We recalculate the ionic strength and iterate several times. m 2 2 when  = 0.7694,  3.32  102 0.125  m m 2  7.0097  103  0.05608 m m 2  0.05608 m  7.0097  103  0 m 0.05608   0.05608 2  4 1  7.0097  103  2  0.06025 mol kg 1 I  0.06025 mol kg 1 ln    1.173  1  0.06025  0.2879    0.7498 when  = 0.7498 m2  0.05905 0.125  m m 2  0.05905 m  7.3815  103  0 m 0.05905   0.05905 2  4 1  7.3815  103  2  0.06132 mol kg 1 I  0.06132 mol kg 1 ln    0.7479    0.7479 when  = 0.7479 m2  0.05935 0.125  m m 2  0.05935 m  7.4191  103  0 m 0.05935   0.05935 2  4 1  7.4191  103  I  0.06143 mol kg 1 ln    0.2907    0.7477 2  0.06143 mol kg 1 when  = 0.7477 m2  0.05938 0.125  m m 2  0.05938 m  7.4228  103  0 m 0.05938   0.05938 2  4 1  7.4228  103  2  0.06144 mol kg 1 This is a sufficiently good result, and we calculate the degree of dissociation to be 0.06144  100%  49%. 0.125 b)  = 1 m 2  4.15  103  3.32  102 m m 2  3.32  102 m  4.15  103  0 m 3.32  102  3.32  10  2 2   4 1 4.15  103 2   0.04992 mol kg 1 0.04992  100%  40% 0.125 This result is 20% smaller than that calculated using the Debye–Hückel limiting law. P9.24) From the data in Table 9.3 (see Appendix B, Data Tables), calculate the activity of the electrolyte in 0.100 m solutions of a. KCl b. H2SO4 c. MgCl2 a) KCl    0.770 0.100 m m  a       m  m  m  m  m2   0.100 0.100 m  0.100 mol kg 1 a   0.100 0.770  0.0770 b) H 2SO 4 0.100 m    0.2655 m3   2  0.100  0.100  4.0  103 2 1 m  0.15874 mol kg 1 a   0.15874   0.2655  0.0422 c) MgCl2 0.100 m    0.529 m3   0.100 2  0.100  4.0  103 2 m  0.15874 mol kg 1 a   0.15874   0.529  0.0840 P9.28) The principal ions of human blood plasma and their molal concentrations are mNa   0.14 m, mCl   0.10 m, mHCO   0.025 m. 3 Calculate the ionic strength of blood plasma. I       1 1 2 mi z i  0.14 m  12  0.10 m  12  0.025 m  12  0.1325 m  2 i 2 P9.29) Estimate the degree of dissociation of a 0.100 m solution of acetic acid (Ka = 1.75  10–5) that is also 0.500 m in the strong electrolyte given in parts (a)–(c). Use the data tables in Appendix B to obtain ±, because the electrolyte concentration is too high to use the Debye–Hückel limiting law. a. Ca(Cl)2 b. KCl c. MgSO4 In each case, the activity of the ions produced in the dissociation of acetic acid is determined by the strong electrolyte. a) Ca  Cl 2  +  1, Cl Ca 2+ 0.500 m 1.00 m   2 z  2, z  1 From Table 10.3,   = 0.448  2 m 2 5 1.75  10  0.100  m m  0.002910 mol kg 1 2 0.448 m 2   0.100  m 0.002910  100%  2.91% 0.100  +  1,    1, z  1, z  1 From Table 10.3,   = 0.649 b) KCl 1.75  105   0.649 2 m2 0.100  m m  2.018  103 mol kg 1 2.018  103  100%  2.02% 0.100  +  1, c) MgSO 4 z  2,   1 z  2 From Table 10.3,   = 0.0675 5 1.75  10   2 m 2 2 0.0675 m 2   0.100  m 0.100  m 1 m  0.01777 mol kg 0.01777  100%  17.7% 0.100 P9.32) Calculate Gr and the equilibrium constant at 298.15K for the reaction         aq  3H 2 g  8H  aq  2Cr 3 aq  7H2 O l . Cr2 O2 7 Using the half cell reactions: Oxidation: 3 H2   6 H   6 e  Reduction: Cr2 O 7 2 E0  0 V  6 e   2 Cr2 3 E 0  1.232 V E cell  E red  E ox  1.232V   ΔG reaction   n F Δφ  6   96485 C mol -1  1.232 V   713.2 kJ mol -1     6  96485 C mol -1 nF  K eq  Exp  E cell   Exp   1.232 V   Exp287.85 -1 -1 R T   8.314472 J mol K  298.15 K     P9.40) By finding appropriate half-cell reactions, calculate the equilibrium constant at 298.15 K for the following reactions: a. 2Cd(OH)2  2Cd + O2 + 2H2O  b. 2MnO2 4  2H 2 O  2MnO 2  4OH  O 2 a) The half cell reactions are Cd(OH)2 + 2e– → Cd + 2 OH– Eº = –0.809V 4OH– → O2 + 2H2O + 4e– Eº = –0.401V The overall reaction is 2 Cd(OH)2 → 2 Cd + O2 + 2H2O Eº = –1.21V nF  4  96485 C mol1 1.21V ln K  E  8.314 J K 1mol1  298.15 K RT  188.391 K  1.52 1082 b) The half cell reactions are MnO42– + 2H2O + 2e– → MnO2 + 4 OH– Eº = + 0.60V 4OH– → O2 + 2H2O + 4e– Eº = –0.401V The overall reaction is 2MnO42– + 2H2O → 2MnO2 +4 OH– + O2 Eº = + 0.20V nF  4  96485 C mol1  0.20 V ln K  E  8.314 J K 1mol1  298.15 K RT  31.139 K  3.34 1013 P9.44) Determine Ksp for AgBr at 298.15 K using the electrochemical cell described by        Ag s AgBr s Br  aq, a Br  Ag aq, a Ag  Ag s . The half cell and overall reactions are AgBr + e– → Ag + Br– Eº = +0.07133 V Ag → Ag+ + e– Eº = –0.7996 V AgBr → Ag+ + Br– Eº = –0.72827 V nE  0.729 V   12.310 0.05916 0.05916 V K sp  4.90  1013 log10 K sp   P9.48) Consider the half-cell reaction O2(g) + 4H+(aq) + 4e–  2H2O. By what factor are n, Q, E, and E° changed if all of the stoichiometric coefficients are multiplied by a factor of 2? Justify your answers. Eo = 1.03 V aO2  1 E  E  RT 1 ln nF  a  4 H a) aH+  0.5 8.3145 J mol K    298.15 K  ln  1 E  1.03 V    2    96, 485 C mol   0.5 1 1 4 1 4    1.0122 V  b) aH+   0.757    0.5  8.3145 J mol K    298.15 K  ln  0.757  0.5       2    96, 485 C mol  1 E  1.03  1 1 4  1.0050 V The relative error is given by 1.0122 V  1.0050 V 100%  0.72% 1.0050 V If rounding off to the correct number of significant figures were done at intermediate stages of the calculation, the relative error would be zero. P9.50) Consider the couple Ox  e    Red with the oxidized and reduced species at unit activity. What must be the value of E° for this half-cell if the reductant R is to liberate hydrogen at 1 atm from a. an acid solution with aH   1 b. water at pH = 7? c. Is hydrogen a better reducing agent in acid or basic solution? a) H+ + e–  ½ H2 Red  Ox + e– E = 0 E  =? Overall: H+ + Red  ½ H2 + Ox Because the activities of H+ and H2 are one, E = E  = 0 V. The same is true of the OxRed couple. To liberate H2, E  for the reaction must be > 0. Therefore, E  for Ox + e–  Red must be < 0. b) For a pH of 7, aH   107 E  E  aH2 RT ln  0.05916 log10 107   0.414 V nF aH  For the overall reaction to be spontaneous, E  for Red  Ox + e– must be > 0.414 V. c) It is a better reducing agent in basic solution, because the oxidation potential for the reaction ½ H2  H+ + e– becomes more positive as the pH decreases P10.1) In anaerobic cells glucose C6H12O6 is converted to lactic acid: C6H12O6  2CH3CH(OH)COOH. The standard enthalpies of formation of glucose and lactic acid are –1274.45 and –694.04 kJ/mol, respectively. The molar heat capacities (at constant pressure) for glucose and lactic acid are 218.86 and 127.6 J/mol K, respectively. a. Calculate the molar enthalpy associated with forming lactic acid from glucose at 298 K. ΔH reaction,298   i ΔH i,reaction,298  products    i ΔH i,reaction,298  reactants  i i   2    694.04 kJ    1   1274.45 kJ   113.64 kJ For one mol of lactic acid: ΔH reaction,298   113.64 kJ mol   56.82 kJ mol 1 2mol  1 b. What would the quantity in part (a) be if the reaction proceeded at a physiological temperature of 310 K? At 310 K the enthalpy is given by: ΔH reaction,310  ΔH i, reaction,298  ΔC p ΔT          56.82 kJ mol -1  2   127.6 J mol -1 K -1   1  218.86 J mol -1 K -1  12 K   56.38 kJ mol -1 c. Based on your answers in parts (a) and (b) how sensitive is the enthalpy of this reaction to moderate temperature changes? Based on the results of a) and b), the enthalpy of reaction is not very sensitive to moderate temperature changes. P10.2) Using the data in Table 10.1, calculate the equilibrium constant for the phosphorylation of fructose by ATP to form fructose-6-phosphate. To use the data in Table 10.1 we have to consider the following pair of reactions: ATP  H 2 O   ADP  Pi G 1  -31.5 kJ mol -1 fructose  Pi   fructose - 6 - phosphate   H 2 O G 2  13.8 mol -1 That means that the standard Gibbs energy for the phosphorylation of fructose is: G phosphorylation  -31.5 kJ mol -1  13.8 mol -1  -16.7 kJ mol -1 The equilibrium constant is then: K phosphorylation    ΔG phosphorylation       16700 J mol -1   Exp   845.6   Exp   -1 -1 R T    8.314472 J mol K    298 K   P10.4) Glutamine (G-ine) is an amino acid formed from the reaction of ammonium ion with glutamate (G-ate–): a. At pH 7 and T = 298 K, the equilibrium constant is 0.003. Calculate the standard Gibbs energy change for the formation of glutamine. b. Glutamine synthesis can be coupled to ATP hydrolysis as follows: Assuming the equilibrium constant for this reaction is 9600, calculate the standard Gibbs energy change. c. From the data in parts (a) and (b), calculate the equilibrium constant for ATP hydrolysis and the Gibbs energy of hydrolysis of ATP. a) The standard Gibbs energy change for the formation of glutamine is given by: ΔG reaction   R T ln  K reaction   8.314472 J mol-1 K -1    298 K   Exp  0.003  14.39 kJ mol-1 b) The standard Gibbs energy change for Glutamine synthesis is given by: ΔG reaction   R T ln  K reaction    8.314472 J mol-1 K -1    298 K   Exp  9600  22.72 kJ mol -1 c) To calculate the equilibrium constant for ATP hydrolysis and the Gibbs energy of hydrolysis of ATP we have to consider the chemical equations from parts a) and b):  G - ine  H 2 O   NH 4  G - ate  NH 4  G - ate -  ATP   G - ine  H 2 O  ADP  Pi G reaction  14.39 kJ mol -1 G reaction  22.72 mol -1 ADP  Pi   ATP G reaction  37.11 kJ mol -1 K hydrolysis   ΔG hydrolysis      37.11 kJ mol -1  6  Exp    Exp    3.20  10 -1 -1   R T        8,314472 J mol K  298 K  P10.5) Calculate the reversible work required to transport 1 mol of sucrose from a region where Csucrose = 0.140 M to a region where Csucrose = 0.340 M. Assume T = 298 K. The reversible work is given by: ~  R T ln  c out   8.314472 J mol -1 K -1   298 K   ln  0.340    2.20 kJ mol -1 Δμ    0.140      c in    P10.6) Calculate the reversible work required to transport 1mol of K+ from a region where to a region where if the potential change accompanying this movement is  = 0.055 V. Assume T = 298 K. The reversible work is given by: c  Δ  R T ln  out   n F   cin    35.5 mM   -1  8.314472 J mol-1 K -1    298K   ln    1   96485 C mol    0.055 V    5.25 mM    10.0 kJ mol -1 P10.7) Suppose the concentration of protons inside a cell membrane is and similarly the concentration outside is . Assume the electrostatic potential difference is  = out – in. a. Obtain an expression for the electrochemical potential difference in terms of the pH difference and the electrostatic potential difference . b. Suppose the pH inside the inner mitochondrial membrane is 0.75 units higher than outside the membrane and  = out – in = –0.168V. Calculate the electrochemical potential difference between protons inside versus outside the membrane. a) We can derive the following expression:       n F Δφ  c  H Δμ  R T ln  out   n F Δφ  R T ln  cout   ln  cin   n F Δφ  R T ln c out  ln cinH  cin   R T ln 10  xΔpH  n F Δφ b) Using the result from part a), the electrochemical potential difference between protons inside versus outside the membrane is calculated as: Δμ  R T ln 10  ΔpH  n F Δφ  8.314472 J mol-1 K -1    298 K   ln 10   0.75  1   96485.3 C mol-1    0.168 V   20.49 kJ mol -1 P10.9) Suppose a membrane is passively permeable to water and to Cl– ion, but not to H+. The electrostatic potential  and the equilibrium concentrations of Cl– and H+ inside and outside the membrane are given here. Assume that the ionic strength inside the membrane is sufficiently small so that the activity coefficients of univalent ions may be taken to be 1 inside the membrane. Assume T = 298 K. Outside Membrane Inside Membrane a. What is the equilibrium concentration of Cl– inside the membrane ? b. Calculate the difference between the chemical potentials of [H+] inside and outside the membrane. c. The reaction is coupled to the transport of H+ with an efficiency of 50% (i.e., half the free energy from ATP hydrolysis is recovered and available to drive the active transport of H+). Assuming the concentration of inorganic phosphate is 0.01 M, what ratio of ATP to ADP is required to drive the transport of H+? a) We are using the equilibrium condition and solve for cin: Δφ  ln R T  c out  ln   zF  cin  c out  zFΔφ   cin  R T  c out  zFΔφ   Exp  cin  R T  cin  c out  zFΔφ  Exp   R T    0.05 M    0.150 V    96485.3 C mol    1  Exp   -1 -1  8.314472 J mol K    298 K   -1  1.45  104 M b) The difference between the chemical potentials of [H+] inside and outside the membrane is given by: c  Δ  R T ln  out   z F Δφ  cin    5  106 M    8.314472 J mol K    298 K   ln     96485.3 C mol -1   1   0.150 V  7  1  10 M    24.16 kJ mol-1 -1 -1 c) To obtain the ratio of ATP to ADP we use the fact that 50% of the energy produced by the hydrolysis of ATP is available: ΔG transport  0.5 ΔG ATPADP  ΔG ATPADP  R T ln  Q   0.01 M   c ADP  2   24.16 kJ mol-1    31 kJ mol -1   8.314472 J mol-1 K -1    298 K   ln   c ATP    0.01 M  c   17.32 kJ mol-1    2477.71 J mol-1   ln   c  ADP  ATP   Exp  7   0.01 M   c ADP c ATP Exp  7 c ADP   0.091  0.01 M  c ATP c ATP  11 c ADP