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CEM 383 Equation Sheet, Final Exam - 2004M = mega = 10 6 k = kilo = 10 3m = milli = 10 -3 µ = micro = 10 -6n = nano = 10 -9 p = pico = 10 -12π = 3.14159261 amu = 1.660 x 10 -27 kgN A = 6.022 x 10 23g = 9.8 m/s 2R = 8.314 m 3 Pa/molKR = 8.314 J/mol KR = 0.08206 Latm/molKk B =1.38 x 10 -23 J/K = R/N Aslope = ∆y/∆xK = o C + 273.15oC = (5/9)( o F–32)oF = (9/5)( o C) + 321 J = 1 kgm 2 /s 2 = 1 Nm1 Pa = 1 kg/ms 2 = 1 Nm -21 atm = 1.01325 x 10 5 Pa1 atm = 760 torr = 760 mm Hg1 bar = 10 3 mbar = 10 5 Paf⎛⎜⎝( s)2n a ⎞+ ⎟V⎠P2= P(s)∆s⎛P(s)= 4π⎜⎝3RTs 2 = =M( V - nb) = nRTm2πkBT⎞⎟⎠32s3k2BmrreT−∆Hln P =vap + const.RTPPAA==xxAAPK∆Tb = Kbm∆Tf=Ptot=P*B+x∗AA12=MM2( −ms2kT )K(* *)P − PABfBm21P = F/Ad = M/VP = nRT/Vd=PM/RTP 1 = (n 1 /n T )P T = x 1 P TE = 1/2 mv 2Ptot=*A*B[* * *(){ }]P − x v P − PAAPPAB

x2∫x1SZdxCEM 383 Equation Sheet, Final Exam - 2004dW = FdsdW = −PdVW = −nRTln( V 2 V1)1 cal = 4.184 J⎛ dQV⎞ ⎛ dE ⎞CV= ⎜ ⎟ = ⎜ ⎟ W = CV ( T 2 −T1)⎝ dT ⎠ ⎝ dT ⎠ VdQ = CdT⎛ dQP⎞ ⎛ dH ⎞CdE = dW + dQP = ⎜ ⎟ = ⎜ ⎟⎝ dT ⎠ ⎝ dT ⎠dZdS = Qrev / TdS = dS +==Z(x,=xtot2= ln Ωk B−xsysy,K)( dZ / dx)dSlimT →0K∆S = CV ln +∆Horxno∆S∆CP=1x2dx∫ = lnxx1surrS=P0( T T ) nR ( V V )2 1 ln∑( x x )212E = K.E.+ UdE = dQ dH =1y , K dx + ( dZ / dy)x,Kdy+n∆HoofV dQ PdE = dH − PdVdE = TdS − PdVdH = TdS + VdP( dT / dV ) = −(dP / dS)S( dT / dP)= ( dV / dS)SL(products) -∑ m∆H= ∑ nS(products) -∑= ∑ nC(products) -∑PmSomCofPV( reactants)( reactants)P( reactants)

∆GCEM383 Equation Sheet, Final Exam - 2004K p = K c (RT) ∆n∆G = ∆G o + RT lnQ∆G o = -RT lnKKxorxn∆G==⎛ Kln⎜⎝ Ke=−∆∑= ∆H− b ±21G o⎞ ∆H⎟ =⎠ Rn∆G/ RTofb2 − 4ac2aorxn(products) -∑ m∆G−T∆S⎛⎜⎝1T1−1T2⎞⎟⎠of( reactants)dG = VdP − SdTdA = −PdV− SdToGi= niGi + niRT ln PioGi= niGi + niRT ln XoG = n Gi + n RT ln Cii( ) idG µµ iµ iµidn i T , P,n j ,K == µoi+ RT ln Piµ i + RT ln Xiµ + RT ln C= o= oiiiii∆G=n∑i=1µ vii∑(products) − µ vni=1ii(reactants)K =pPν iiνP ii(prod)(reac)Pν jjν jjP(prod) L(reac) LKc=[i(prod)][i(reac)]ννii[j(prod)][j(reac)]ννjjLL

CEM 383 Equation Sheet, Final Exam - 2004ln [A] t = −kt+ ln[A] 0 K w = [H + ][OH - ]K w =1.0 x 10d[A]-14arate = − = k[A]pH = - log [H + ]dtpOH = -log[OH - ]pH = 14.0 - pOH−K w = K a × K b[ ] [ ]ktA A ett 1/2 =1[A]t1/2= 00.693k⎛ kln⎜⎝ k12⎞⎟⎠= −ER1= 2kt+t [A] k = Ae -E a / RT0=12k[A]0Z=a⎛⎜⎝1T1−1T2⎟ ⎞⎠2πNAσPvRT[[]t[ A] 0 k1( ) ( −kt k te e )1 −−k − kB 2t== [ A]2⎧⎨1+⎩11( ) ( −kt −ktk e − k e )12 1k − kC 2]012⎫⎬⎭K =kkfr

ECEM383 Equation Sheet, Final Exam - 20041 V = 1J / C 1 A = 1C/ s1 F = 96,485C / mole=1e∗∆GF== 1.602 × 10=Fq= −nFEAqqA B204πεr=qB4πεr02−19dG = VdP − SdT + ,φ(r)=qBe4πεεr20⎛⎜ 8πNAeρκ =⎜⎝1000⋅4πε0εklim∆φ =i→0E=EE0 −solventBE o∆GCdW rev other−κrT=⎞⎟⎟⎠00.5ε96,485J / Vmole0= 8.854×10−12C2/ JmqAqBU =∗E = iR l4πε 0 rU qBφ = = I = 0.5* ∑ iq 4πε riI= −nFE0.5RTln KnF0.0592logQnln0A 0A =∆φ = −iRV = iRm zµ i = µ0i + RT ln aimaii = γn±i⎛ m ⎞0a = ⎜±⎟ m± γ ± 0⎝ m ⎠lim γ =1n± = n++ n−m →ii 0(n)n+ n 1/−±γ ± = γ + ⋅γ−mγ ±(n) + n 1/ n−±m m± = + ⋅−= A⋅z+⋅ z−( 2πNρ )Asolvent⋅ I1/ 20.5⎛ 2⎜ e⎜⎝4πε0εkb2i⎞⎟T ⎟⎠3/ 2

CEM383 Equation Sheet, Final Exam - 2004S= k B ln Ωln N!≈ N ln NWqE==N!∏n∑Nj=1n∑j=1=n∑j=1n iigegQ = qQ=jqjjgee!jNNN!−ee−ej−j−ejNkkBTBTkBTN=n∑j=1n j∑j=1E−einigiePi= =nNg ej=k−en∑jj=1BTkBTn j e jn i gi−∆Ek T= eB , ∆E= ein j g−jEHG=NkBT2⎛⎜⎝ed ln q ⎞⎟dT ⎠⎛ d ln q ⎞= NkBT⎜T+ 1⎟⎝ dT ⎠jV⎛ q ⎞= −NkBT⎜ln −1⎟⎝ N ⎠µ = −k T lniBqni