More Tutorial at www.littledubdoctor.co Physical Cheistry Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILY understand your work; only work on the exa sheet will be considered. Write answers, where appropriate, with reasonable nubers of significant figures. You ay use only the "Student Handbook," a calculator, and a straight edge. 1. (10 Points) Coplete the sentence in the left colun using the answers provided in the right colun. Where requested, explicitly write out True or False so as to avoid abiguity. DO NOT WRITE IN THIS SPACE 1. The following expression applies generally to binary azeotropes, y i P = x i P saturation i. True or False? False 2. The Third Law of Therodynaics arises fro the observation that as absolute teperature approaches zero, the F of a pure fluid also ust approach zero to ensure that we can obtain a nonsingular value of entropy at T = 0 Kelvin. 3. In the Van der Waals equation of state, the O paraeter relates to the attractive interactions between particles. 4. For a pure fluid, the nuber of intensive degrees of freedo along the solid-vapor equilibriu curve is H. 5.The Henry s Law ideal fugacity for a species in solution is C. 6. Consider the solidification of ice; the entropy change of the ice is negative. What can we say about the entropy change of the environent? E 7. Debye-Huckel theory treats a solvent as M. 8. In ulti-coponent systes, N relates individual species activities to one another. 9.The Henry s Law activity coefficient, γ HL i, approaches unity as the concentration of species i approaches L. 10. Defining the equilibriu constant in ters of reactant and product concentrations is the ost general forulation of this therodynaic quantity. True or False? FALSE a. reversible b. d c. ˆ f i liquid = k i HL x i d. entropy decreases e. entropy increases f. heat capacity g. 2 + C - P h. 1 i. ixing j. b k. entropy is constant due to reversible heat exchange l. zero. a continuous, atoically unresolved ediu of constant dielectric n. Gibbs-Duhe equation o. a p. 2 p. 1 /10 p. 2 /10 p. 3 /10 p. 4 /10 p. 5 /10 p. 6 /10 p. 7 /10 p. 8 /10 p. 9 /10 p. 10 /10 ============= p.11 /10 (Extra credit) ============= TOTAL PTS /100
NAME: 2 2. (10 Points) Consider the binary syste coposed of acetonitrile(1) and nitroethane(2). This syste fors ideal vapor and liquid ixtures. At therodynaic conditions where the liquid ixture is in equilibriu with the gas ixture: A. How any independent intensive variables (degrees of freedo) are available? F = 2 + C P = 2 + 2 2 = 2 i.e. (P and {x}) or (T and {x}) or ({x} and {y}) B. At a total pressure, P Total = 70 kpa, and liquid coposition of x 1 = 0.5156, what are the teperature and vapor phase coposition, again considering that vapor and liquid phases are in equilibriu. The following inforation ay be helpful. Teperature in Celsius and pressure in kilopascal (kpa) for the following equations. ln P 1 saturation =14.2724 2945.47 T +224.0 ln P 2 saturation =14.2043 2972.64 T +209.0 Solution: This question asks for the teperature at the given pressure and liquid-phase coposition. This will be an iterative solution, since the teperature enters non-linearly into the analysis via the relations for the saturation vapor pressures given above. Keep in ind that the vapor is an ideal gas ixture and the solution is ideal, thus following Raoult s expression.
More Tutorial at www.littledubdoctor.co NAME: 3 P Total = P 1 + P 2 = y 1 P Total + y 2 P Total = x 1 P saturation saturation 1 + x 2 P 2 saturation P Total = P saturation P 2 x 1 1 saturation P + x 2 (A) 2 Define, α 12 α 12 = P saturation 1 P ln saturation ( α 12) = 0.0681 2945.47 2 T +224.0 + 2972.64 T +209.0 (B) P Total P saturation 2 = = saturation P x 1 1 saturation P + x 2 2 P Total ( ) x 1 α 12 + x 2 (C) Iteration procedure: 1. Pick initial teperature 2. Copute α 12 (equation B) saturation 3. Copute P 2 (equation C) 4. Copute T using inverted Antoine relation 2972.64 T = 14.2043 ln P 209.0 saturation 2 5. Copute new α 12 and P 2 saturation fro equations B and C 6. Iterate until convergence Results of iterations Iteration # Tep for step 2 Tep for step 4 Difference between Tep for steps 2 and 4 1 77.4065030730214 70.0000000000000 7.40650307302144 2 77.9577905146540 77.4065030730214 0.551287441632553 3 77.9970941220114 77.9577905146540 3.930360735745353E-002 4 77.9998874331253 77.9970941220114 2.793311113862273E-003 5 78.0000859095134 77.9998874331253 1.984763880500395E-004 6 78.0001000118627 78.0000859095134 1.410234932563981E-005 We achieve convergence within 6 iterations even with a bad guess of 70 Celsius in the initial step. Vapor phase coposition:
More Tutorial at www.dublittledoctor.co NAME: 4 y 1 P Total = x 1 P 1 saturation y 1 = x 1P 1 saturation P Total = = (0.007366)(91.743) = 0.6758 ( 0.5156) e 70 2945.47 14.2724 78+224
NAME: 5 3. What is the solubility of AgCl in 0.01 olal NaCl solution? Consider the reaction as: Solution AgCl(solid) Ag + (aqueous) + Cl (aqueous) This is a proble asking for the equilibriu olality of the ions in solution. We write the foral definition for this process: K = e ΔG o / RT = a a Ag + Cl = a a Ag + a Cl = γ Ag + Ag +γ Cl Cl = γ 2 ± Ag + Cl AgCl e ΔG o / RT = γ ± 2 Ag + Cl Thus, we need to copute the standard free energy of reaction, the ean activity coefficient, and the equilibriu olality expressions for this syste. First, the standard free energy of reaction. o o o o ΔG rxn = ΔG foration,ag + + ΔG foration,cl ΔG foration,agcl = 77.1 kj ol kj + 131.2 109.8 kj = 55.7 kj ol ol ol = 55.7x103 J ol Thus J 55.7x10 3 K = exp ol J (8.3145155.7 ol - K ) ( 298.15K) =1.76x10 10 Now deterine ean activity coefficient via Debye-Huckel relation.
NAME: 6 I ln γ ± = α DH Z + Z 1+ Ba o I take Ba o 1 (as discussed in class) I ln γ ± = α DH Z + Z 1+ I I = 1 2 i Z 2 1 i i = 2 ( x + 0.01+ x + 0.01) = x + 0.01 ( ) Thus ( x + 0.01) ( 1.177) ( x + 0.01) ln γ ± = α DH Z + Z = 1+ ( x + 0.01) 1+ ( x + 0.01) ( 1.177) ( x + 0.01) γ ± = exp 1+ ( x + 0.01) ( 1.177)(2) ( x + 0.01) γ 2 ± = exp 1+ ( x + 0.01) x=2.18x10-8 ol/kg γ ± =0.899
More Tutorial at www.dublittledoctor.co NAME: 7 4. Much effort has been directed toward finding drugs to cobat the AIDS syndroe caused by HIV. Therodynaics akes a contribution to that effort by facilitating interpretation of experiental calorietric titration data. Isotheral titration calorietry (ITC) provides inforation about the energetics of the interactions between physiologically iportant acroolecules and sall substrates (i.e. low-olecular weight drugs noncovalently interacting with, i.e. binding to, protein receptor targets). An exaple is the inhibitor drug Ritonavir binding to a polypeptide cleaving site of HIV-1 protease (a protease is a protein that cleaves peptide bonds) shown in Figure 1. (The binding of the drug in the binding site prevents the cleavage of the viral polypeptide chain into functionally active proteins the virus needs to propagate, Figure 1, right panel). Figure 1. (left) HIV-1 protease and a sall-olecule inhibitor. (Right) The HIV viral life cycle showing the junction where inhibition of the protease is targeted in order to shut down viral propagation. The ITC ethod is based on easuring the increental aounts of heat requiring reoval fro or addition to a solution of the acroolecule as sall aounts of the ligand (i.e., the sall olecule) are added over tie. In the siplest practical scenario, the reversible binding reaction is pictured as: Protein (P) + Ligand (L) K B Protein : Ligand (PL) By assuing that there is a single binding site, one can derive relationships between the enthalpy of binding at standard conditions ( ΔH o ), the binding constant, K B, and the nuber of binding sites (in this case taken to be 1 binding site). For the HIV-1 protease syste, binding constants fro ITC easureents at 25 Celsius are shown in Table 1. The enthalpies of binding, fro the sae set of experients, are also shown. Deterine the associated changes in Gibbs free energy and entropy for the various drugs. Ritonavir Saquinavir Nelfinavir Indinavir H O (kcal/ol) -3.70 1.90 2.60 2.10 K B 22.65 19.97 19.30 20.13 G O (kcal/ol) -1.85-1.78-1.76-1.778 S O (kcal/ol-k) -0.0062 0.012 0.015 0.013 Binding enthalpy data fro: Luque et al. PROTEINS: Structure, Function, Genetics. 49:181-190. 2002. Binding constant data fro: Valezquez-Capoy et al. Archives of Biocheistry and Biophysics. 390:169-175. 2001.
NAME: 8 Solution: K = e ΔG o / RT ΔG o = RT lnk ΔS o = ΔH o ΔG o T More Tutorial at www.littledubdoctor.co T=298K, R=0.00199 kcal/ol K RT= 0.593 kcal/ol
NAME: 9 5. Consider the expansion process of Heliu gas initially at 27 Celsius and pressure = 2 x 10 5 N/ 2. I. The gas first undergoes constant pressure expansion to a volue of 1.5 3. II. Process A is followed by a reversible adiabatic expansion to a final state of volue = 2.0 3 and a pressure one-half of the initial pressure. A. How uch work is done during the expansion in step I? V 2 w = p ext dv = p 1 V 2 V 1 V1 ( ) = p 1 V 2 nrt 1 P 1 B. What is the change in internal energy for step I? Ideal gas, internal energy only depends on teperature ΔU = nc V ΔT = nc V (T 2 T 1 ) C. What is the heat exchanged for step I? q = ΔU w = nc V (T 2 T 1 ) + p 1 V 2 nrt 1 D. What is the entropy change for step II? P 1 Reversible and adiabatic, thus: δq ΔS = rev = 0 T
More Tutorial at www.littledubdoctor.co NAME: 10 6. For the following process at T = -10 Celsius and P = 1 bar (constant T and P) in which water spontaneously and copletely freezes, derive a relation for the entropy change for water. H 2 O(l,T = 10 o C,P =1bar) H 2 O(s,T = 10 o C,P =1bar)
More Tutorial at www.littledubdoctor.co NAME: 11
NAME: 12 7. Consider a rubber band of length L aintained at a tension f. The total differential of the internal energy of the rubber band coprised of n oles of aterial is: du = T ds + f dl + µ dn where U is the internal energy, T is the teperature, S is the entropy, f is the tension, L is the length, µ is the cheical potential of the rubber band, and n is the nuber of oles of aterial. The equation of state (EOS) of this syste is U = θs 2 L. Derive the Gibbs-Duhe relation for this syste. n 2 Solution: du = T ds + f dl + µ dn Thus U = TS + fl + µn Gibbs - Duhe relation is derived siply by subtracting the canonical pairs: 0 = U TS fl µn Taking the dotal differential of this last expression gives: 0 = du T ds S dt f dl L df µ dn n dµ 0 = S dt L df n dµ 0 = S dt + L df + n dµ The last two expressions are the G-D relation for this syste.
NAME: 13 8. For a second order phase transition, the olar enthalpy, entropy, and volue for a pure fluid are continuous functions through the phase change. Thus, like the equality of cheical potentials of a pure fluid in the coexisting phases for a first-order transition, one can write the equality of the olar volue, enthalpy, and entropy of the two phases (denote the α and β) along the coexistence line for such fluids. Consider the equality of olar volue for this proble. Based on the above inforation, deterine a relation analogous to the Clausius-Clapeyron equation that relates the slope along the coexistence line, dp to the coefficient of dt theral expansion and isotheral copressibility of the coexisting phases of the fluid. Recall that olar volue, a state function, can be expressed in ters of state variables such as teperature and pressure; that is, V = V (T,P). Thus, its total differential can be expressed as: Solution V α = V β dv (T,P) = V (T,P) T Moving along the coexistence line, the above equality leads to : dv α β = dv Thus V α (T,P) dt + V α (T,P) dp = V β (T,P) T P P T T But V α τ α V dt V α κ α V dp = V β τ β V dt V β κ β V dp where the coefficient of theral expansion is τ V α = 1 κ V α = 1 V α V α P Thus, since V α = V β T τ α V dt κ α V dp = τ β V dt κ β V dp κ α β ( V κ V )dp τ α β ( V τ V )dt =1 dp dt = τ α β ( V τ V ) κ α β V κ V ( ) P P dt + V (T,P) dp P dt + V β (T,P) dp P V α α V T P T T and the isotheral copressibility is
More Tutorial at www.littledubdoctor.co NAME: 14 9. Consider the following recently postulated pressure-teperature phase diagra for a fluid. The phase diagra suggests the following: 0 < dp dt subliation < dp dt fusion < dp dt vaporization This data indicates two things. First, there is an unusual property of this fluid; what is it? Second, this fluid violates a fundaental law of therodynaics. What law is this? Support your answer with appropriate discussion. Think about the forulation of the Clapeyron equation along the subliation and vaporization lines in ters of relative agnitudes of pure fluid olar entropies and olar volues to help guide your thinking about this question. Solution: A. The property that is unusual is the slope of the solid-liquid equilibriu line that is negative; ost fluids do not have a negative slope, but rather a positive one. Water is one exaple of a fluid with a negative slope. B. The Clapeyron equation gives: dp dt dp dt vaporization subliation = S gas liquid S V gas liquid V = S gas solid S V gas solid V S gas liquid S gas V S gas solid S gas V Since the postulated phase diagra suggests that the vaporization slope is greater than the subliation slope, this would have to ean that:
NAME: 15 S gas S liquid V gas > S gas solid S gas V iplies S liquid solid < S Since the olar entropy of liquids is higher than that of solids, this violates the 2 nd law of therodynaics!
NAME: 16 10. An ideal Carnot engine, with an efficiency η = 0.40, operates using 0.5 kilooles of an ideal diatoic gas as the working substance. During the isotheral expansion stage, the pressure of the gas decreases to half of the axiu pressure on the cycle. At the end of the adiabatic expansion stage, the pressure of the gas is 9 at and its volue is 2 3. A. Calculate the heat absorbed fro the high teperature reservoir during the isotheral expansion stage of the cycle. B. Calculate the work done by the gas during each of the 4 stages of the cycle.
NAME: 17 C. Calculate the entropy changes for each of the 4 stages of the cycle.
NAME: 18 11 (Extra Credit 10 Points). The total differential of the internal energy of a rubber band coprised of n oles of aterial is: du = T ds + f dl + µ dn where U is the internal energy, T is the teperature, S is the entropy, f is the tension, L is the length, µ is the cheical potential of the rubber band, and n is the nuber of oles of aterial. The equation of state (EOS) of this syste is U = θs 2 L. Show that this equation of state satisfies the Gibbs-Duhe relation for this syste. Solution: n 2 More Tutorial at www.littledubdoctor.co 0 = S dt + L df + n dµ Gibbs - Duhe equation Fro the total differential of the internal energy, we know the following (recall that this equation is a fundaental equation for this syste and gives us inforation) T = U S L,n = 2θSL n 2 ; f = U L S,n = θs 2 n ; µ = U 2 n L,S = 2θS 2 L n 3 Now copute the total derivatives required for the Gibbs- Duhe equation : dt = T S L,n ds + T L S,n dl + T n L,S dt = 2θL ds + 2θS dl + 4θSL dn n 2 n 2 n 3 dn df = f ds + f dn S n n df = 2θS ds + 2θS 2 dn n 2 n 3 S dµ = µ S n,l ds + µ L n,s dl + µ n S,L dn dµ = 4θSL ds + 2θS 2 dl + 6θS 2 L dn n 3 n 3 n 4 Putting the total differentials for T, f, and µ into the Gibbs - Duhe relation, we find that all ters annihilate each other leaving the desired result, identical equality to 0.