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1 CHEM 3411 MWF 9:00AM Spring 2012 Physical Chemistry I Final Exam, Version A (Dated: May 4, 2012 Name: GT-ID: NOTE: Partial Credit will be awarded! However, full credit will be awarded only if the correct answer is in marked boxes, when provided. (Do not write in the small boxes in the right margin; they are for grading purposes only. You MUST sign the honor pledge: Part Problem Max.Pts Actual Score I / II / III / IV / TOTAL: 120 On my honor, I pledge that I will not give or receive aid in examinations; I will not attempt to gain prior access to examinations; I will not represent the work of another as my own; and I will avoid any activity which will encourage others to violate their own pledge of honor. Your Signature: You may find the following information useful: Constant Symbol Value Units Speed of light in vacuum c E+10 cm sec 1 Electronic charge e E 10 esu e E 19 C (Coulombs Avogadro s number N A E+23 molecules mole 1 Gas Constant R J K 1 mole 1 R E 2 L bar K 1 mole 1 R E 2 L atm K 1 mole 1 R E+1 L Torr K 1 mole 1 Boltzmann constant k B E 23 J, K 1 Electron mass m e E 4 amu Proton mass m p amu Neutron mass m n amu Planck Constant h E 27 erg sec h E 34 J sec h E 27 erg sec Bohr radius a E-11 m Atomic mass unit amu E 24 g Electron volt ev E 19 J ev E 12 erg Debye D E 30 C m Calorie cal J Rydberg Constant R E+5 cm 1 A sheet of possibly useful formulas is provided at the end. 1

2 THIS PAGE INTENTIONALLY LEFT BLANK 2

3 Part I: Thermodynamics: First Law 1. Concept Questions. Try to be concise, and correct (5 pts a. What is the equation of state for a gas described by the virial expansion? Provide a qualitative explanation for how this equation corrects for the ideal gas equation. (5 pts b. What is the first law of thermodynamics, and how does it allow you to define internal energy? 3 :

4 2. First Law: The machinery for gases. A sample consisting of 3.3 moles of ideal gas molecules with C P,m = 19.5JK 1 mol 1 is initially at 162 kpa and 323 K. It then undergoes a reversible adiabatic expansion until its temperature is 247 K. (5 pts a. What is the initial volume? (5 pts b. What is q for this process? Answer: Answer: 4

5 (5 pts c. What is the final pressure? (4 pts d. What is w for this process? Answer: Answer: 5 :

6 Part II: Thermodynamics: Second & Third Law 3. Concept Questions. Try to be concise, and correct (5 pts a. What does the phase rule tell you about the phase transition of a typical pure liquid to a solid? Indicate this on a phase diagram being careful to label your axes. (5 pts b. What is the second law of thermodynamics, which thermodynamic variable(s does it help define, and how? 6 :

7 4. Heat Engines. Consider a heat engine that operates between 780.0K and 340.0K. (5 pts a. What is the maximum efficiency of the engine? Answer: (5 pts b. Calculate the maximum work that can be done by each 1.0kJ of heat supplied by the hot source? Answer: 7

8 (5 pts c. How much heat is discharged into the cold sink in a reversible process for each 1.0kJ supplied by the hot source? Answer: (5 pts d. What is the change in entropy in a reversible process for each 1.0kJ supplied by the hot source? Answer: 8 :

9 Part III: The Thermodynamics of Chemical Systems 5. Concept Questions. Try to be concise, and correct (5 pts a. Suppose you have a nonideal mixture of two components, A and B, which satisfies the regular solution model. Let ǫ αβ be the interaction energy for the contact of component α with component β. If 1 2 (ǫ AA +ǫ BB < ǫ AB, draw a sketch of the excess Gibb s free energy of mixing as a function of the composition of component A? (Make sure to label the axes! (5 pts b. Suppose that you find that electrons are flowing from the solution in beaker A to beaker B. What can you say about the chemical potentials in both beakers? What else must also be happening between the two beakers to enable this flow? 9 :

10 6. Reactions. Suppose that a dissociating reaction A(g 2B(g + C(g takes place in a closed container with fixed pressure equal to 1.00 bar. The reaction is described by the chemical equilibrium expression, ln K = A + B/T + C/T 2 where A = 1.13, B = 932 K and C = K 2, at all temperatures between 300K and 500K. (5 pts a. What is the standard entropy of this reaction at K? Answer: (5 pts b. What is the standard enthalpy of this reaction at K? Answer: 10

11 (5 pts c. Suppose that the system is ideal, and initiated with 1.0 moles of A at constant temperature, K. Estimate the chemical composition of A at equilibrium. Answer: (5 pts d. Suppose that the system is nonideal, and initiated with 1.0 moles of A at constant temperature, K. The activity coefficient of components A and B are 1.00 and the activity coefficient of component C is equal to 1.11 times its mole fraction. Estimate the chemical composition of A at equilibrium. Answer: 11 :

12 Part IV: Kinetics & Beyond 7. Concept Questions. Try to be concise, and correct (5 pts a. Suppose that you have a sequential reaction process, A B C. If the rate of the first step is much faster than that for the second step, what can you say about how the concentration of B and C with time? (5 pts b. What does the prefactor in the transition state theory rate forumula have to do with the underlying assumptions? 12 :

13 8. Arrehnius Rates (10 pts a. The rate constant for the decomposition of a certain subtance is dm 3 mol 1 s 1 at 300.0K and dm 3 mol 1 s 1 at 330.0K. Whate are the Arrhenius parameters of the reaction? Answer: Answer: 13

14 (5 pts b. How much would you need to lower the activation energy of this reaction in order to double the rate constant? Answer: (5 pts c. Suppose, instead that a bimolecular gas reaction takes place at constant volume. How do you expect that the prefactor in the Arhenius rate will change (if at all if the temperature is doubled? Justify. 14 :

15 (Possibly Useful Formulas WARNING: Use at your own risk PV = nrt PV m = RTZ x a = na n T ( ( Z = 1 + BP + CP 2 + Z = 1 + B 1 + C V m P = nrt a ( n 2 1 Z = (a/rt V nb V 1 (b/v m V m 2 1 V m + U = q + w du = dq + dw H = q P U = w ad U = q V ( Vf V w = P ex V i dv w = P ex V w = nrt ln f V i H = U + PV P = ( U V S,N ( α 1 V V T V = ( H P S,N du = π T dv + C V dt du = PdV + TdS P ( dh = (µc P dp + C P dt κ T 1 V dh = V dp + TdS V P T C V = ( U C T V P = ( H C T P P C V = nr C V = 3nR 2 V T c = constant c = C V,m R PV γ = constant γ = 1 + R C V,m i νr i R i j νp j P j r H = j νp j f H (P j i νr i f H (R i ds = dqrev T S = T 2 T 1 C p T dt r C P,m = j νp j C P,m (P j i νr i C P,m (R i S vap = Hvap T b 85 J mol S surr = Hsys T ( ǫ w q h = 1 Tc T h A = U TS G = U + PV TS da = dw max dg = dw add,max r S m = j νp j S m(p j i νr i S m(r i r G = j νp j f G (P j i νr i f G (R i ds dq 0 ds T U,V 0 du S,V 0 ds H,P 0 dh S,P 0 da T,V 0 dg T,P 0 ( ( G ( = H P G(P T T T P 2 f P i = V P G(P f P i = NRT ln f P i G(P = G + NRT ln ( f f = φp ln φ = P Z 1 dp P 0 P dp = trss dt trsv µ = G N = G m dµ = S m dt + V m dp p = p e ( Vm P dp dt = trshm T trsv m ln ( ( P P = vaph m ( 1 1 R T T RT ( [ P = P + fus H m fus ln ( ] T V T dw = γdσ P in = P out + 2γ r h = 2γ ρgr cos θ c = γsg γ sl γ lg w ad = γ sg + γ lg γ sl Continued on NEXT Page 15

16 V j = ( V n j P,T,n µ j = ( G n j G = P,T,n i n dµ iµ B i dµ A = n A nb dg = V dp SdT + i µ idn i P A = x A PA P B = x B K B Π = ( n B V RT µ = µ + RT ln ( P (P µ P A (l = µ A (l + RT ln A µ PA A (l = µ A (l + RT ln (x A mix G = nrt(x A ln x A + x B ln x B + βx A x B mix S = nr(x A ln x A + x B ln x B mix G = mix G (id + nrt(x A ln γ A + x B ln γ B µ = µ + RT ln(a a A = P A PA ( b a = γx a = γ B PA B b y A = x A x B = vaph R ( r G G ξ dln(k dt ( T T 2 P,T = rh RT 2 PB +(P A P B x A ( T T 2 T = K b b x B = fush R r G = j νp j µ(p j i νr i µ(r i ln(k 2 /K 1 = rh R ( 1 T 2 1 T 1 a B = P B K B n a l( ad = n b l( db F = C P + 2 T = K f b Q P j P a j νp j P i R a i νr i ln x B = fush R ( 1 1 T T RT ln(k = r G ( solv G = z2 i e2 N A 8πǫ 0 r i 1 1 ǫ r ( µ = µ id + RT ln γ γ ± = (γ + γ 1 2 log(γ ± = Z + Z AI 1 2 A = I = 1 2 i z2 bi i b I k ( b i k(x,m + = 1 k(x,m 2+ = 3 k(x 2,M + = 3 k(x 2,M 2+ = 4 b c 2 = 1 3 v2 ( Z = σ c N rel V r G = νfe E = E RT ln Q F = N νf Ae v 2 e Mv2 /(2RT P( v = ( 3 M 2 e Mv2 /(2RT f(v = 4π ( M 2πRT 2πRT λ c Z λ = kt 2σP Z w = ( N crel V 4 ( T z P = ( 2πRT M 1/2 m A 0 t κ = 1νk 3 Bλ cn κ effusion = Z w A 0 D = 3 cλ 1 κ E = 1λ cc 3 V,M[A] η = 1Nmλτ 3 J z = κ ( T J z z = D ( ( N J = κ T z E J z z = η ( v x z η u η e Ea/(RT G = 1/R κ = (GL/A Λ m κ/c λ ion κ/c ion Λ m = ν + λ + + ν λ Λ m = Λ m κc 1 2 κ = A + BΛ M Λ m = αλ m 1 Λ m = 1 Λ M + ΛmC k a(λ m 2 K a = α2 C 1 α F E = ZeE = Ze φ/l f F = 6πηa s = (Ze/f F E u s/e = (Ze/(6πηa λ = ZuF f F = ( µ = ( C x P,T RT J = D ( C s = Df F D = urt C x P,T x P,T RT Zf F Λ m = ( ν + Z 2 +D + + ν Z 2 D F 2 RT D = kt f F C = J t x C = D 2 C t x 2 C = D 2 C ν C kc t x 2 x C(x,t = n 0 A(πDt 1/2 e x2 /(4Dt C(r,t = n 0 e r2 /(4Dt x 2 1 8(πDt 3/2 2 = (2Dt 1 2 x = st D = λ2 2τ ν(t = [A] t ν(t = 1 [A] a t = + 1 c [C] t ν(t = k[a] n ν(t = k[a] m [B] n k 2 = σ ( 1 8k B T 2 πµ k 2 = ( RT P k K [A] = [A] 0 e akt t 1/2 = ln 2 k = c rel σ K = ( k B T hν k 2 = ( ( kt RT h p NA q C q A q B 1 1 [A] [A] 0 = akt k = Ae E A/(RT e E/(RT k = κν k 2 = σ N A ce E/(RT e S/R e H/(RT 16 k 2 = kak d k a+k d k d = 4πR D

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