FinalExamSolutions2003

FinalExamSolutions2003 - Final Exam / Chemistry 4511/...

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Unformatted text preview: Final Exam / Chemistry 4511/ Physical Chemistry for Engineers Helpful Information: PV= nRT 760 TolT= 1 atm 1 bar = 750 TOlT R = 0.08206 atm dm3K-Imorl R = 8.3145 J K-I morl kB= 1.381 x 10-23J K-I Avogadro's Number = 6.022 x 1023morl dm = O.lm dm3= 1000cm3= 1 Liter 1.0 A= 1.0 X 10-10m 1 J = 1 kg m-2 S-2 1 amu = 1.66 x 10-27kg 1 amu x Avogadro's Number = Mass of mole in grams ( ) 1/2 2 / ZA = 2 ndA <uA>NAV ZAA= 0.5 ndA2 <uA>NA2N2 <UA> = (8RT/nM)I/2= (8kT/nm)1/2 A = V/[(2)1/2n dA2NA] dN/N = 4n (m/2nkT)3/2 exp[-mu2I2kT] U2du f1U= q (heat absorbed by system) + w (work done on system) f1U = q - Pf1V H=U+PV f1H= f1U+ f1(PV) f1(PV)= f1nRT qp = f1H f1U= 0 for isothermal expansion w = nRT In (VIN2) for work done on the system for isothermal reversible process f1S = q/T Isothermal Expansion: f1S= nRln(V 2N I) Efficiency = 1- Te/T h G=H-TS f1G = f1H - Tf1S f1G < 0 is spontaneous f1Go = -RT In Kpor f1Go= -RT In Kc (Different standard states for these two f1Goequations) t1/2 = 1n2/k A(t) = A(O)exp(-kt) t1/2= l/[A(O)k] l/A(t)- l/A(O) = kt Steady State Approximation, d[I]/dt = 0 k = A exp[-E/RT] k = (kBT/h) exp[-~~Go/RT] ~~Go = ~~Ho- T~~So First-order unimolecular gas reaction: First-order unimolecular gas reaction: e = 2.718 Ea = ~~Ho + RT k = e(kBT/h)exp[~~So/R]exp[-Ea/RT] aZ + bz = CZ nA= 2d sinE> Cubic Lattice: dhkl = a/(hz+ kz + f)1/Z E>= K[A]/(l + K[A]) (Langmuir Adsorption Isotherm) (llNa)[P/(Po-P)] = [(c-1)/(c A no)](P/Po)+ (lI(c A no) (BET Adsorption Isoth Various terms in BET Adsorption Isotherm: Na = Total amount adsorbed Po= Saturation Pressure c = 11hz no = Number of sites per cmz A = Total surface area (cmz) 1 eV = 1.602 x 10-19J C=YA c = 3.0 X 108 m S-I ~E = hY h = 6.626 X 10-34Js hY = (ll2)mvZ + <1> A= hip p=mv Energy operator = i(hl2n)8Iat Momentum rerator = -i(hl2n)8Iax E = nzhz/8ma mass of electron, m=9.109 x 10-31kg T=IIIo A = -log IIIo A = EC t exp[-i2nEt/h] = exp[-icot] 2ny = co "---.-. EJ = J(J+1)hz/8nzI ~E = 2(J+1)hz/8nzI For J-+ J+ 1 Rotational Transitions: v = 2(1+ 1)h/8nzI = 2(J+ 1)B B =h/8nzI v"'= 1/A= vie = 2(J+ 1)h/8nzIc = 2(1+ 1)B'" B'" =h/8nzIe 1- z- flro fl = mlmz/(ml+mz) Ev= (v+1/2) hvo For v -+ v+1 Vibrational Transistions: ~E = hvo Vo= (1/2n)(k/fl)l/2 G(v) = vo(v + 1/2) -Xe(v+ 1/2)2 ~G(v-+v+ 1) = ~Gv+1/Z= vo[1-2Xe(v+1)] Do = 2:~Gv+1/Z So I Vt h \(J n /(e,- F/~~ J E xaYVi I BASICS & GASES '----- 1. ' ' . Energy attempts to distribute itself in as random a way as possible. Based on the example in class, assume that you have 4 particles that can distribute themselves into four different boxes with energies E= 0,1,2,3 and 4. The total energy, ET,for the 4 particles is fixed at E-r=3. The particles are not distinguishable within a specific box. However, particles are distinguishable if they are in different boxes. Particles can also exchange between the various boxes. What arrangement of particles in the various boxes is the most random and yields the largest number of distinguishable combinations?...
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This note was uploaded on 02/26/2008 for the course CHEM 4521 taught by Professor Bierbaum during the Spring '03 term at Colorado.

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FinalExamSolutions2003 - Final Exam / Chemistry 4511/...

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