Q4 (2 marks) The limiting molar conductivities are 9.10, 42.6, and 12.65 mS mmol for CH3COONa, HCI, NaCl, respectively. Calculate the limiting molar conductivity for CH;COOH. Given 1 AG=A,G + RT In Q AG =- RT INK -- In A, H1 R TT K log yt=-12-2|41%, 1=2, z (b,/b) , A=0.509 for solutions of two types of ions at molalities b, and b 1 = 2(b. 2? + b_2?)/b A,G --VFE Nernst equation E=E" RT In Q VF at 25C E=E 0.0257 v In AG=AH-TA,S at constant T and P (OG/AT)p=-5 number density N = N/V; N: number of molecules , V: volume N= NNA M = MNA; M: molar mass, m: molecular mass PV = nRT REK NA 3/2 The Maxwell-Boltzmann distribution of speeds: f(v) = 41 Kohlrausch law: Am = A - K and A = v. 2.+v-1 Am: molar conductivity, AM: limiting molar conductivity, K: constant, c: molar concentration, V+ and v- : number of cations and anions per formula unit of electrolyte, v?e-Mv2/2RT 2nRT 1 and 1- : limiting molar conductivity of cations and anions. Am = k/c ,x: conductivity sa u E, u= ze / 6 ana , E = A/! 1 S: drift speed, u: ion mobility, Eclectric field, ze charge of ion, n: viscosity, a: radius of ion , Aq: potential difference between electrodes, ;distance between electrodes conductivity and mobility 1 = z uF Arrhenius equation k= A et/RT k: rate constant, A: pre-exponential factor, Ea : activation energy order of reaction A Products integrated rate law 0 [A] = [A]. + kt 1 In TAL =-kt 2 +kt [AL 4RE * = aE * : induced dipole moment; a: polarizability; E: electric field Polarizability volume o: permittivity of vacuum Lennard-Jones potential V=48{@)"-3)} &: depth of the well; re : separation at which V = 0;r:intermolecular distance Coulomb potential energy in vacuum: V = 9:07 ANET Taylor series expansions for x