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1 Chemistry 3411 MWF 9:00AM Spring 2009 Physical Chemistry I Final Exam, Version A (Dated: April 30, 2009 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
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3 Part I: Thermodynamics: First Law 1. Concept Questions. Try to be concise, and correct! (5 pts a. Provide a chemical example showing that heat is not a state function. (5 pts b. What is the principle of corresponding states, and to what extent is it relevant to ideal or nonideal gases? S: S: 3 :
4 2. Heat & Work. A mixture of 2.45g of helium and 6.35g of argon initially at C is allowed to expand reversibly and adiabatically from 543 cm 3 to 5.43 dm 3. Assume that both gases and their mixture are ideal. (5 pts a. What is the initial pressure of the sample? (5 pts b. What is the final pressure of the sample? 4
5 (5 pts c. What is the final temperature of the sample? (5 pts d. What is the work for this reversible adiabatic path? 5 :
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7 Part II: Thermodynamics: Second & Third Law 3. Concept Questions. Try to be concise, and correct! (5 pts a. What is a spontaneous process? What are the thermodynamic variable(s that govern when processes are spontaneous? And how? (5 pts b. How is the Helmholtz energy related to internal energy, and what type of experiment would Helmholtz energy be useful for? S: S: 7 :
8 4. Fugacity and Coexistence Curves (5 pts a. What is the fugacity of 89.23mg of N 2 at 485.0K in a 1.000L container if the gas is treated as perfect? (5 pts b. What is the fugacity of 89.23mg of N 2 at 485.0K in a 1.000L container if the molar Gibbs energy of the actual gas were to be.834 kj/mol less than the molar Gibbs energy of the corresponding perfect gas? 8
9 For parts c and d, an unknown one-component system exhibits the following points on the phase diagram: T = 73.40K and P = 55.21kPa at the triple point, T = K and P = kPa at the critical point, a solid-liquid coexistence point at T = K and P = kPa, and a solid-gas coexistence point at T = 11.85K and P = 1.142kPa, (5 pts c. What is the formula describing the coexistence curve between solid and gas on a P-T phase diagram? (Make sure to provide values for all coefficients. 9
10 (5 pts d. What is the formula describing the coexistence curve between liquid and gas on a P-T phase diagram? (Make sure to provide values for all coefficients. 10 :
11 Part III: The Thermodynamics of Chemical Systems 5. Concept Questions. Try to be concise, and correct! (5 pts a. Suppose that the unimolecular isomerization, A B, is exothermic. What will happen to the equilibrium concentrations of the isomer if the temperature is increased? Justify. (5 pts b. Suppose that you have a mixture with roughly equal amounts of a surfactant (such as soap, water and oil. How many phases will this mixture give rise to, and what is the approximate composition of each phase? Is your answer consistent with Gibbs phase rule? Justify! S: S: 11 :
12 6. For this problem, consider the following gas-phase reaction: 2A + 2B 2C + 3D (1 (8 pts a. After mixing 1.15 mol A, 1.00 mol B, and.850 mol C, the resulting equilibrium mixture contained.175 mol D at 298 K and 1.00 bar. Calculate the mole fraction of each species at equilibrium. x A : S: x B : S: x C : S: x D : S: 12
13 (4 pts b. Suppose, instead, that the system described by the chemical equation provided above is at equilibrium at 298 K and 1.00 bar when it consists of.795 mol A,.893 mol B,.578 mol C, and.435 mol D? Use this information to calculate the equilibrium constant at 298 K and 1.00 bar? (4 pts c. What is r G for the system of part b? 13
14 (4 pts d. What is mix S for the gases with respect to the equilibrium mole fractions of part b? 14 :
15 Part IV: Kinetics & Beyond 7. Concept Questions. Try to be concise, and correct! (5 pts a. What are the key assumptions of the kinetic model? (5 pts b. How does convection play a role in describing the concentration profile of a solute in a liquid? S: S: 15 :
16 8. Kinetics You may assume that the diameter of water is m (5 pts a. What is the root mean square speed of the molecules in a water vapor at 1891K and pressure of atm? (5 pts b. What is collision frequency between molecules in a water vapor at 1891K and pressure of atm? 16
17 (5 pts c. Given that the mobility of Li + in water is m 2 s 1 V 1, what is the diffusion constant of Li + in water at room temperature? (5 pts d. Calculate the hydrodynamic radius of CH 3 OH in water given that its diffusion constant is m 2 s 1 at room temperature. (Assume that water has a viscosity of kg m 1 s 1 : 17
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19 (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 w = P ex Vf V i dv H = U + PV du = π T dv + C V dt dh = (µc P dp + C P dt C V = ( U T V P = ( U V α 1 V C P = ( H T P w = P ex V ( V T κ T 1 V S,N P ( V P T V = ( H P S,N du = PdV + TdS dh = V dp + TdS C P C V = nr C V = 3 2 nr V T 1/c = constant c = C V,m PV γ = constant γ = 1 + R 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 19
20 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 + βx A 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 rmbλ cn κ effusion = Z w A 0 D = 1 3 cλ 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 NA q C qa q B p 1 1 [A] [A] 0 = akt k = Ae E A/(RT e E/(RT k = κν k 2 = σ N A c 1 2e E/(RT e S/R e H/(RT 20 k 2 = kak d k a+k d k d = 4πR D
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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
You MUST sign the honor pledge:
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