Spring 2011 MT3 Nitsche Solutions

Spring 2011 MT3 Nitsche Solutions - leyChemistry 1A, Spring...

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Page 1 of 15 leyChemistry 1A, Spring 2011 Midterm 3 April 11, 2011 (90 min, closed book) Name:___________________ SID:_____________________ TA Name:________________ There are 40 multiple choice questions worth 3 points each. There is only one correct answer for each question unless otherwise specified. Only answers on the Scantron form will be graded. Scantron must be properly filled in and cannot contain any smudges or other marks. Scantrons will not be rescanned! You can tear off the equation sheet and the periodic table for your convenience. You can use the page margin or the back of the pages as scratch paper. You can take the exam booklet with you after the exam.
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Page 2 of 15 Thermodynamics: G ° = H ° - T S ° H ° = Σ H ° f (products) - Σ H ° f (reactants) S ° = Σ S ° (products) - Σ S ° (reactants) G ° = Σ G ° f (products) - Σ G ° f (reactants) S = k B lnW S = q rev /T E = q + w w = - P ext V for aA + bB cC + dD b a d c B A D C Q ] [ ] [ ] [ ] [ = At equilibrium, Q = K G = G ° + RTln Q G = G° + RTln(a); a = activity = γP/P° or γ[A]/[A]° G ° = - RTln K G ° = - nF Є º Є = Є º - (RT/nF) lnQ R S T R H K ° + ° = 1 ln T = ik b,f m Π = iMRT P total = P A + P B = X A P A ° + X B P B ° q=cm(T 2 -T 1 ) Kinetics: [A] t = [A] 0 e -kt ln[A] t = ln[A] 0 – kt t 1/2 = ln2/k 1/[A] t = 1/[A] 0 + kt k = A e (-Ea/RT) ln(k 1 /k 2 ) = E a /R ( 1/T 2 – 1/T 1 ) t 1/2 = 1/[A] 0 k t 1/2 = [A] 0 /kt Quantum: E = h ν λν = c λ deBroglie = h / p = h / mv E kin (e-) = h ν - Φ = h ν - h ν 0 = R n Z E n 2 2 x ∆p ~ h p = mv E n = h 2 n 2 /8mL 2 ; n = 1, 2, 3. .. E v = (v + ½) hA/2π; A =(k/m) ½ E n = n(n + 1 ) hB; B = h/8π 2 I; I = 2mr 2 m = m A m B /(m A + m B ) Ideal Gas: PV = nRT RT E kin 2 3 = M 3RT v rms = Constants: N 0 = 6.02214 x 10 23 mol -1 R = 2.179874 x 10 -18 J R = 3.28984 x 10 15 Hz k = 1.38066 x 10 -23 J K -1 h = 6.62608 x 10 -34 J s m e = 9.101939 x 10 -31 kg c = 2.99792 x 10 8 m s -1 T (K) = T (C) + 273.15 F = 96,485 C / mol 1 V = 1 J / C Gas Constant: R = 8.31451 J K -1 mol -1 R = 8.20578 x 10 -2 L atm K -1 mol -1 1 nm = 10 -9 m 1 kJ = 1000 J 1 atm = 760 mm Hg = 760 torr ≈ 1 bar 1 L atm 100 J
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Page 3 of 15 1. One of the underlying assumptions of the kinetic theory of gases is A. Gas molecules move randomly at fixed velocities. B. Gas molecules only exert forces when they move close to one another. C. Energy is lost only during collisions with the container. D. All of these are assumptions of the kinetic theory of gases. E. None of these is an assumption of the kinetic theory of gases. ANS: E 2. At 25°C we have the following root-mean-square speeds (in m/s): Xe - 238, Ar - 431, He - 1360. Thus sulfur dioxide gas molecules at 25°C should have a root-mean-square speed A. < 238 m/s. B. > 238 m/s but < 431 m/s.
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This note was uploaded on 05/06/2011 for the course CHEM 1A taught by Professor Nitsche during the Spring '08 term at University of California, Berkeley.

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Spring 2011 MT3 Nitsche Solutions - leyChemistry 1A, Spring...

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