MT1_v4.2-1

MT1_v4.2-1 - Chemistry 1A, Spring 2009 Midterm 1, Version...

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Page 1 of 11 Chemistry 1A, Spring 2009 Midterm 1, Version February 9, 2009 (90 min, closed book) Name:___________________ SID:_____________________ TA Name:________________ x There are 22 Multiple choice questions worth 3 points each. x There are 3, multi-part short answer questions. x For the multiple choice section, fill in the Scantron form AND circle your answer on the exam. x Put your written answers in the boxes provided. Full credit cannot be gained for answers outside the boxes provided. x The lecture, homework, chemquizzes, discussion or experiment that each question is based upon is listed after the question e.g. [L3, HW 1.13, CQ 7.3] Question Points Score Multiple Choice Section 66 Question 23 12 Question 24 9 Question 25 13 Total 100
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Page 2 of 11 Thermodynamics: ' G q = ' H q - T ' S q ' H q = 6 ' H q f (products) - 6 ' H q f (reactants) ' S q = 6 S q (products) - 6 S q (reactants) ' G q = 6 ' G q f (products) - 6 ' G q 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 q + RTln Q G = G° + RTln(a); a = activity = Ȗ P/P° or Ȗ [A]/[A]° ' G q = - RTln K ' G q = - nF ' ȯ º ' ȯ = ' ȯ º - (RT/nF) lnQ R S T R H K q ' ± q ' ² 1 ln ¨ T = ik b,f m Ȇ = iMRT P total = P A + P B = X A P A ° + X B P B ° Acid Base: pH = - log[H 3 O + ] pX = - log X ] [ ] [ log HA A pK pH a ² ± 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 Q OQ = c O deBroglie = h / p = h / mv E kin (e-) = h Q - ) = h Q - h Q 0 f ² R n Z E n 2 2 ¨ x ¨ p ~ h p = mv Particle in a box (1-D Quantum): E n = h 2 n 2 /8mL 2 ; n = 1, 2, 3. .. Vibrational: E v = (v + ½) hA/2 ʌ ; A =(k/m) ½ Rotational: E n = n(n + 1) hB; B = h/8 ʌ 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 f = 2.179874 x 10 -18 J R f = 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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P a g e 3 o f 1 1 M ULTIPLE C HOICE The illustration to below represents a mixture of nitrogen (black) and oxygen (grey) molecules. [Disc1]
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This note was uploaded on 05/04/2011 for the course CHEM 1A taught by Professor Nitsche during the Spring '08 term at Berkeley.

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MT1_v4.2-1 - Chemistry 1A, Spring 2009 Midterm 1, Version...

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