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 exam sheet will be considered. Write answers, where appropriate, with reasonable numbers of significant figures. You may use only the "Student Handbook," a calculator, and a straight edge. 1. (10 Points) Complete the sentence in the left column using the answers provided in the right column. Where requested, explicitly write out True or False so as to avoid ambiguity.

DO NOT WRITE IN THIS SPACE 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

1. The following expression applies generally to binary azeotropes,

€

yi P = xi Pisaturation . True or False?

_______False_____

a. reversible

2. The Third Law of Thermodynamics arises from the observation that as absolute temperature approaches zero, the ________F______ of a pure fluid also must approach zero to ensure that we can obtain a non-singular value of entropy at T = 0 Kelvin.

b. d

3. In the Van der Waals equation of state, the _____O___ parameter relates to the attractive interactions between particles.

c.

€

ˆ f iliquid = ki

HL xi

4. For a pure fluid, the number of intensive degrees of freedom along the solid-vapor equilibrium curve is _____H_____.

d. entropy decreases

5.The Henry’s Law ideal fugacity for a species in solution is _______C_______.

e. entropy increases

6. Consider the solidification of ice; the entropy change of the ice is negative. What can we say about the entropy change of the environment? ______E_____

f. heat capacity

7. Debye-Huckel theory treats a solvent as _____M____.

g. 2 + C - P

8. In multi-component systems, ___N__ relates individual species activities to one another.

h. 1

9.The Henry’s Law activity coefficient,

€

γ iHL ,

approaches unity as the concentration of species “i” approaches ___L___.

i. mixing

10. Defining the equilibrium constant in terms of reactant and product concentrations is the most general formulation of this thermodynamic quantity. True or False? ______FALSE____

j. b

k. entropy is constant due to reversible heat exchange

l. zero m. a continuous, atomically

unresolved medium of constant dielectric

n. Gibbs-Duhem equation o. a p. 2

Physical Chemistry

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2. (10 Points) Consider the binary system composed of acetonitrile(1) and nitromethane(2). This system forms ideal vapor and liquid mixtures.

At thermodynamic conditions where the liquid mixture is in equilibrium with the gas mixture: A. How many independent intensive variables (degrees of freedom) 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, PTotal = 70 kPa, and liquid composition of x1 = 0.5156, what are the temperature and vapor phase composition, again considering that vapor and liquid phases are in equilibrium. The following information may be helpful. Temperature in Celsius and pressure in kilopascal (kPa) for the following equations.

€

ln P1saturation =14.2724 − 2945.47

T +224.0

ln P2saturation =14.2043− 2972.64

T +209.0

Solution: This question asks for the temperature at the given pressure and liquid-phase composition. This will be an iterative solution, since the temperature enters non-linearly into the analysis via the relations for the saturation vapor pressures given above. Keep in mind that the vapor is an ideal gas mixture and the solution is ideal, thus following Raoult’s expression.

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€

PTotal = P1 + P2 = y1PTotal + y2PTotal = x1P1saturation + x2P2

saturation

PTotal = P2saturation x1

P1saturation

P2saturation + x2

⎛

⎝ ⎜

⎞

⎠ ⎟ (A)

Define, α12

α12 =P1saturation

P2saturation ⇒ ln α12( ) = 0.0681− 2945.47

T +224.0+

2972.64T +209.0

(B)

P2saturation =

PTotal

x1P1saturation

P2saturation + x2

⎛

⎝ ⎜

⎞

⎠ ⎟

=PTotal

x1α12 + x2( ) (C)

Iteration procedure: 1. Pick initial temperature 2. Compute

€

α12 (equation B) 3. Compute

€

P2saturation (equation C)

4. Compute T using inverted Antoine relation

€

T =2972.64

14.2043− ln P2saturation − 209.0

5. Compute new

€

α12 and

€

P2saturation from equations B and C

6. Iterate until convergence Results of iterations Iteration # Temp for step 2 Temp for step 4 Difference between

Temp 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 composition:

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€

y1PTotal = x1P1saturation

y1 =x1P1

saturation

PTotal=

0.5156( ) e14.2724− 2945.47

78+224⎛

⎝ ⎜

⎞

⎠ ⎟

70= (0.007366)(91.743)= 0.6758

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NAME: 5

3. What is the solubility of AgCl in 0.01 molal NaCl solution? Consider the reaction as:

€

AgCl(solid)→Ag+(aqueous) +Cl−(aqueous)

Solution This is a problem asking for the equilibrium molality of the ions in solution. We write the formal definition for this process:

€

K = e−ΔGo /RT =

aAg +aCl −aAgCl

= aAg +aCl − = γAg +mAg +γCl −mCl − = γ±2mAg +mCl −

e−ΔGo /RT = γ±

2mAg +mCl −

Thus, we need to compute the standard free energy of reaction, the mean activity coefficient, and the equilibrium molality expressions for this system. First, the standard free energy of reaction.

€

ΔGrxno = ΔGformation,Ag+

o + ΔGformation,Cl−o − ΔGformation,AgCl

o

= 77.1 kJmol

+ −131.2 kJmol

⎛

⎝ ⎜

⎞

⎠ ⎟ − −109.8 kJ

mol⎛

⎝ ⎜

⎞

⎠ ⎟ = 55.7 kJ

mol= 55.7x103 J

mol

Thus

K = exp −55.7x103 J

mol(8.3145155.7 J

mol - K) 298.15K( )

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

=1.76x10−10

Now determine mean activity coefficient via Debye-Huckel relation.

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NAME: 6

€

ln γ± = −αDH Z+Z−I

1+ Bao I

⎛

⎝ ⎜

⎞

⎠ ⎟

take Bao ≈1 (as discussed in class)

ln γ± = −αDH Z+Z−I

1+ I

⎛

⎝ ⎜

⎞

⎠ ⎟

I =12

Zi2mi =

i∑ 1

2x + 0.01+ x + 0.01( ) = x + 0.01( )

Thus

ln γ± = −αDH Z+Z−x + 0.01( )

1+ x + 0.01( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ =

(−1.177) x + 0.01( )1+ x + 0.01( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

γ± = exp(−1.177) x + 0.01( )

1+ x + 0.01( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

γ±2 = exp

(−1.177)(2) x + 0.01( )1+ x + 0.01( )

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

x=2.18x10-8 mol/kg γ± =0.899

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NAME: 7

4. Much effort has been directed toward finding drugs to combat the AIDS syndrome caused by HIV. Thermodynamics makes a contribution to that effort by facilitating interpretation of experimental calorimetric titration data. Isothermal titration calorimetry (ITC) provides information about the energetics of the interactions between physiologically important macromolecules and small substrates (i.e. low-molecular weight drugs non-covalently interacting with, i.e. binding to, protein receptor targets). An example 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).

The ITC method is based on measuring the incremental amounts of heat requiring removal from or addition to a solution of the macromolecule as small amounts of the ligand (i.e., the small molecule) are added over time. In the simplest practical scenario, the reversible binding “reaction” is pictured as:

€

Protein (P) + Ligand (L) KB← → ⎯ Protein : Ligand (PL) By assuming that there is a single binding site, one can derive relationships between the enthalpy of binding at standard conditions (

€

ΔHo ), the binding constant, KB, and the number of binding sites (in this case taken to be 1 binding site). For the HIV-1 protease system, binding constants from ITC measurements at 25 Celsius are shown in Table 1. The enthalpies of binding, from the same set of experiments, are also shown. Determine the associated changes in Gibbs free energy and entropy for the various drugs. Ritonavir Saquinavir Nelfinavir Indinavir ∆HO (kcal/mol) -3.70 1.90 2.60 2.10 KB 22.65 19.97 19.30 20.13 ∆GO (kcal/mol) -1.85 -1.78 -1.76 -1.778 ∆SO (kcal/mol-K) -0.0062 0.012 0.015 0.013 Binding enthalpy data from: Luque et al. PROTEINS: Structure, Function, Genetics. 49:181-190. 2002. Binding constant data from: Valezquez-Campoy et al. Archives of Biochemistry and Biophysics. 390:169-175. 2001.

Figure 1. (left) HIV-1 protease and a small-molecule inhibitor. (Right) The HIV viral life cycle showing the junction where inhibition of the protease is targeted in order to shut down viral propagation.

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Solution:

€

K = e−ΔGo /RT

ΔGo = −RT lnK

ΔSo =ΔHo −ΔGo

T

T=298K, R=0.00199 kcal/mol K RT= 0.593 kcal/mol

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5. Consider the expansion process of Helium gas initially at 27 Celsius and pressure = 2 x 105 N/m2. I. The gas first undergoes constant pressure expansion to a volume of 1.5 m3. II. Process A is followed by a reversible adiabatic expansion to a final state of volume = 2.0 m3 and a pressure one-half of the initial pressure. A. How much work is done during the expansion in step I?

€

w = − pextdVV1

V 2

∫ = −p1 V2 −V1( ) = −p1 V2 −nRT1P1

⎛

⎝ ⎜

⎞

⎠ ⎟

B. What is the change in internal energy for step I? Ideal gas, internal energy only depends on temperature

€

ΔU = nCVΔT = nCV (T2 −T1) C. What is the heat exchanged for step I?

€

q = ΔU − w = nCV (T2 −T1) + p1 V2 −nRT1P1

⎛

⎝ ⎜

⎞

⎠ ⎟

D. What is the entropy change for step II? Reversible and adiabatic, thus:

€

ΔS =δqrevT∫ = 0

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6. For the following process at T = -10 Celsius and P = 1 bar (constant T and P) in which water spontaneously and completely freezes, derive a relation for the entropy change for water.

€

H2O(l,T = −10oC,P =1bar)→H2O(s,T = −10oC,P =1bar)

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7. Consider a rubber band of length L maintained at a tension f. The total differential of the internal energy of the rubber band comprised of n moles of material is:

€

dU = T dS + f dL + µ dn

where U is the internal energy, T is the temperature, S is the entropy, f is the tension, L is the length,

€

µ is the chemical potential of the rubber band, and n is the number of moles of material. The equation of state (EOS) of

this system is

€

U =θS2Ln2

. Derive the Gibbs-Duhem relation for this system.

Solution:

€

dU = T dS + f dL + µ dnThusU = TS + fL + µnGibbs - Duhem relation is derived simply by subtracting the canonical pairs :0 =U −TS − fL −µnTaking 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 system.

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NAME: 13

8. For a second order phase transition, the molar enthalpy, entropy, and volume for a pure fluid are continuous functions through the phase change. Thus, like the equality of chemical potentials of a pure fluid in the coexisting phases for a first-order transition, one can write the equality of the molar volume, enthalpy, and entropy of the two phases (denote them

€

α and β) along the coexistence line for such fluids. Consider the equality of molar volume for this problem. Based on the above information, determine a relation analogous to

the Clausius-Clapeyron equation that relates the slope along the coexistence line,

€

dPdT

to the coefficient of

thermal expansion and isothermal compressibility of the coexisting phases of the fluid. Recall that molar volume, a state function, can be expressed in terms of state variables such as temperature and pressure; that is, Vm = Vm(T,P). Thus, its total differential can be expressed as:

€

dVm (T,P) =∂Vm (T,P)

∂T⎛

⎝ ⎜

⎞

⎠ ⎟ P

dT +∂Vm (T,P)

∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

dP

Solution

€

Vmα =Vm

β

Moving along the coexistence line, the above equality leads to :dVm

α = dVmβ

Thus

∂Vmα (T,P)∂T

⎛

⎝ ⎜

⎞

⎠ ⎟ P

dT +∂Vm

α (T,P)∂P

⎛

⎝ ⎜

⎞

⎠ ⎟ T

dP =∂Vm

β (T,P)∂T

⎛

⎝ ⎜

⎞

⎠ ⎟ P

dT +∂Vm

β (T,P)∂P

⎛

⎝ ⎜

⎞

⎠ ⎟ T

dP

ButVm

ατVα dT −Vm

ακVα dP =Vm

βτVβ dT −Vm

βκVβ dP

where the coefficient of thermal expansion is

€

τVα =

1Vm

α

∂Vmα

∂T⎛

⎝ ⎜

⎞

⎠ ⎟ P

and the isothermal compressibility is

€

κVα =

1Vm

α

∂Vmα

∂P⎛

⎝ ⎜

⎞

⎠ ⎟ T

Thus, since

€

Vmα =Vm

β

€

τVα dT −κV

α dP = τVβ dT −κV

β dP

κVα −κV

β( )dPτVα − τV

β( )dT=1

dPdT

=τVα − τV

β( )κVα −κV

β( )

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9. Consider the following recently postulated pressure-temperature phase diagram for a fluid.

The phase diagram suggests the following:

€

0 < dPdT⎛

⎝ ⎜

⎞

⎠ ⎟

sublimation< −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 fundamental law of thermodynamics. What law is this? Support your answer with appropriate discussion. Think about the formulation of the Clapeyron equation along the sublimation and vaporization lines in terms of relative magnitudes of pure fluid molar entropies and molar volumes to help guide your thinking about this question. Solution: A. The property that is unusual is the slope of the solid-liquid equilibrium line that is negative; most fluids do not have a negative slope, but rather a positive one. Water is one example of a fluid with a negative slope. B. The Clapeyron equation gives:

€

dPdT⎛

⎝ ⎜

⎞

⎠ ⎟

vaporization

=Smgas − Sm

liquid

Vmgas −Vm

liquid ≈Smgas − Sm

liquid

Vmgas

dPdT⎛

⎝ ⎜

⎞

⎠ ⎟

sublimation

=Smgas − Sm

solid

Vmgas −Vm

solid ≈Smgas − Sm

solid

Vmgas

Since the postulated phase diagram suggests that the vaporization slope is greater than the sublimation slope, this would have to mean that:

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€

Smgas − Sm

liquid

Vmgas >

Smgas − Sm

solid

Vmgas

impliesSmliquid < Sm

solid

Since the molar entropy of liquids is higher than that of solids, this violates the 2nd law of thermodynamics!

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NAME: 16

10. An ideal Carnot engine, with an efficiency η = 0.40, operates using 0.5 kilomoles of an ideal diatomic gas as the working substance. During the isothermal expansion stage, the pressure of the gas decreases to half of the maximum pressure on the cycle. At the end of the adiabatic expansion stage, the pressure of the gas is 9 atm and its volume is 2 m3. A. Calculate the heat absorbed from the high temperature reservoir during the isothermal expansion stage of the cycle.

B. Calculate the work done by the gas during each of the 4 stages of the cycle.

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NAME: 17

C. Calculate the entropy changes for each of the 4 stages of the cycle.

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NAME: 18

11 (Extra Credit 10 Points). The total differential of the internal energy of a rubber band comprised of n moles of material is:

€

dU = T dS + f dL + µ dn

where U is the internal energy, T is the temperature, S is the entropy, f is the tension, L is the length,

€

µ is the chemical potential of the rubber band, and n is the number of moles of material. The equation of state (EOS) of

this system is

€

U =θS2Ln2

. Show that this equation of state satisfies the Gibbs-Duhem relation for this system.

Solution:

€

0 = S dT + L df + n dµ ← Gibbs - Duhem equationFrom the total differential of the internal energy, we know the following (recall that this equation is a fundamental equation for this system and gives us information)

T =∂U∂S⎛

⎝ ⎜

⎞

⎠ ⎟ L,n

=2θSLn2 ; f =

∂U∂L⎛

⎝ ⎜

⎞

⎠ ⎟ S,n

=θS2

n2 ; µ =∂U∂n⎛

⎝ ⎜

⎞

⎠ ⎟ L,S

=−2θS2Ln3

Now compute the total derivatives required for the Gibbs- Duhem equation :

dT =∂T∂S⎛

⎝ ⎜

⎞

⎠ ⎟ L,n

dS +∂T∂L⎛

⎝ ⎜

⎞

⎠ ⎟ S,n

dL +∂T∂n⎛

⎝ ⎜

⎞

⎠ ⎟ L ,S

dn

dT =2θLn2

⎛

⎝ ⎜

⎞

⎠ ⎟ dS +

2θSn2

⎛

⎝ ⎜

⎞

⎠ ⎟ dL +

−4θSLn3

⎛

⎝ ⎜

⎞

⎠ ⎟ dn

df =∂f∂S⎛

⎝ ⎜

⎞

⎠ ⎟ n

dS +∂f∂n⎛

⎝ ⎜

⎞

⎠ ⎟ S

dn

df =2θSn2

⎛

⎝ ⎜

⎞

⎠ ⎟ dS +

−2θS2

n3

⎛

⎝ ⎜

⎞

⎠ ⎟ dn

dµ =∂µ∂S⎛

⎝ ⎜

⎞

⎠ ⎟ n,LdS +

∂µ∂L⎛

⎝ ⎜

⎞

⎠ ⎟ n,SdL +

∂µ∂n⎛

⎝ ⎜

⎞

⎠ ⎟ S,Ldn

dµ =−4θSLn3

⎛

⎝ ⎜

⎞

⎠ ⎟ dS +

−2θS2

n3

⎛

⎝ ⎜

⎞

⎠ ⎟ dL +

6θS2Ln4

⎛

⎝ ⎜

⎞

⎠ ⎟ dn

Putting the total differentials for T, f, and µ into the Gibbs - Duhem relation, we find that all terms annihilate each other leaving the desired result, identical equality to 0.

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