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Unformatted text preview: CEM 152  Spring 2008 – Exam 4
M = mega = 106 k = kilo = 103 m = milli = 103 µ = micro = 106 n = nano = 109 p = pico = 1012 1Å = 108 cm 3 RT µ= M r1 M2 = r2 M1 π = 3.1415926
1 amu = 1.660 x 1027 kg NA = 6.022 x 1023 g = 9.8 m/s2 R = 8.314 m3Pa/molK R = 8.314 J/molK R = 0.08206 Latm/molK slope = ∆y/∆x K = oC + 273.15 oC = (5/9)(oF – 32) oF = (9/5)(oC) + 32 1 Pa = 1 kg/ms2 = 1 Nm2 1 atm = 1.01325 x 105 Pa 1 atm = 760 torr = 760 mm Hg 1 bar = 103 mbar = 105 Pa P = F/A P = hρg P = nRT/V d = M/V d=PM/RT P1 = (n1/nT)PT = X1PT average K.E. = 1/2 mµ2 n 2a P + 2 (V  nb ) = nRT V ln P =
s= q m × ∆T  ∆H vap RT +C ∆H = H products − H reactants
Sg = kPg PA = XAPoA ∆Tb = Kbm ∆Tf = Kfm π = (n/V)RT = MRT ln [A]t = − kt + ln[A]0
1 1 = kt + [A]t [A]0
t 1/2 = 0.693 k t 1/2 = 1 k[A]0 k = Ae Ea / RT
k E 1 1 ln 1 = a − k 2 R T2 T1 CEM 152  Spring 2008 – Exam 4 [P]p [Q]q k f Kc = = a b [A] [B] kr
Kp = (PP )p (PQ )q (PA )a (PB )b K p = K c ( RT ) ∆n ∆So = ∑ nSo ( products)  ∑ mSo (reactants) ∆H o = ∑ n∆H o (products)  ∑ m∆H o (reactants) rxn f f Kw = [H+][OH]=1.0 x 1014 pH =  log [H+] pOH = log[OH] pH = 14.0 – pOH pKa = log Ka − b ± b 2 − 4ac x= Kw = Ka × Kb 2a ∆G o = ∑ nG o (products)  ∑ mG o (reactants) f f
Eocell = Eo(cathode)  Eo(anode) k = 1.38 x 1023 J/K Eo = Eored(reduction)  Eored(oxidation) 1 N = 1 kgm/s2 1 J = 1 kgm2/s2 ∆Go = –nFEo; ∆G = –nFE 1 cal = 4.184 J 1 F = 96,500 C/mol e∆E = q + w G = H  TS E = Eo  (0.0592/n) logQ ∆H = qp ∆G = ∆H  T∆S logK = nEo/0.0592 ∆E=qv o + RT lnQ G = ∆G wmax = ∆G = nFE qrxn = C ∆T ∆Go = RT lnK S = k ln W K = e − ∆G ∆S = qrev/T ∆S= nR ln (V2/V1) ∆Suniv = ∆Ssys + ∆Ssurr
o / RT 1 V = 1 J/C 1 A = 1 C/s 1 W = 1 J/s 1 kWhr = 3.6 x 106 J ...
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 Spring '08
 MANTICA

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