Cheat Sheet for Exam #1 on Thermodynamics and ... - Chemistry

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ElectrochemistryElectronegativities∆G = -nFE C 2.5 H 2.1∆G = ∆G 0 + RTlnQ S 2.5 Cu 1.9E =E 0 - (0.0592/n) log 10 Q F 4.0 O 3.5log 10 K eq = nE 0 /0.0592 N 3.0W elec = q EZUMDAHL TABLE 11.1 HEREUseful constants and conversions: 1 cal = 4.184 J 1 lit-atm = 101.3 JF = 96,485 coul/mole e = 1.6 x 10 -19 coul/electron 1 Debye = 3.336 x 10 -30 coul-mc = 3 x 10 8 m/s k B = 1.38 x 10 -23 J/K R = 8.31 J/mole-KN 0 = 6.02 x 10 23 molecules/mole 1 amu = 1.66 x 10 -27 kg h = 6.63 x 10 -34 J-sε 0 = 8.85 x 10 -12 coul 2 /J-m m electron = 9.1 x 10 -31 kg 1 Ampere = 1 coul/sec1 nm = 10 -9 m 1 Angstrom = 10 -10 mAverage bond energies (KJ/mole)Atomic Radii (pm)N-H 391 C-N 305 N=O 607 Cs 265C≡C 839 C=N 615 N≡N 941 F 72H-H 432 I-I 149 C=C 614 C 91N-N 160 C-C 347 O=O 495 Cl 99O-H 467 C-O 358 C=O 799
Quantum Mechanicsphoton energy: Heisenberg Uncertainty: DeBroglie Wavelength:E = hν = hc/λ ∆p∆x ≥ h/4π λ = h/mvSchrödinger Equation Hψ = Eψ : (-h 2 /8π 2 m)d 2 ψ/dψ 2 + V(x)ψ = EψH-atom: V = -Ze 2 /4πε 0 r E n = -Z 2 me 4 /(8n 2 h 2 ε 0 2 ) = 2.178 x 10 -18 (Z 2 /n 2 ) (in Joules) a 0 = 0.0529 nmSAMPLE HYDROGENICWAVEFUNCTIONS HERE.Dipole moment: µ = qRMelting Points and Heats of Fusion: Intermolecular potentialMelting Point Enthalpy of Fusion (KJ/mol) (Lennard-Jones)O 2 -218 0.45 V = - aR -6 + bR -12HCl -114 1.99HI -51 2.87CCl 4 -23 2.51CHCl 3 -64 9.2NaF 992 29.3NaCl 801 30.2ln ((P vap (T 1 ) /( P vap (T 2 )) = (∆H vap /R)(1/T 2 - 1/T 1 ) Clausius-Clayperon EquationP = k H χ Henry’s Law: k H = 413 atm (CH 4 in water), 1640 atm (CO 2 in water), 43400 atm (O 2 in water)P(solution) = χ(solvent)P 0 (solvent) : Raoult’s Law
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Cheat Sheet for Exam #1 on Thermodynamics and ... - Chemistry

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