equation08-4

equation08-4 - CEM 152 Spring 2008 – Exam 4 M = mega =...

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Unformatted text preview: CEM 152 - Spring 2008 – Exam 4 M = mega = 106 k = kilo = 103 m = milli = 10-3 µ = micro = 10-6 n = nano = 10-9 p = pico = 10-12 1Å = 10-8 cm 3 RT µ= M r1 M2 = r2 M1 π = 3.1415926 1 amu = 1.660 x 10-27 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 Nm-2 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 10-14 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 10-23 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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equation08-4 - CEM 152 Spring 2008 – Exam 4 M = mega =...

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