Calculate the lattice energy of AgCl(s) using the following thermodynamic data (all data is in kJ/mol). Note that the data given has been perturbed, so looking up the answer is probably not a good idea. Ag(s) AHsublimation = 265 kJ/mol Ag(g) Ionization energy = 711 kJ/mol Cl-CI(g) Bond energy = 223 kJ/mol CI(g) Electron affinity = -369 kJ/mol AgCl(s) AH°r=-147 kJ/mol kJ/mol Do you expect this value to be larger or smaller than the lattice energy of AgBr(s)?

Calculate the lattice energy of RbH (s) using the following thermodynamic data (all data is in kJ/mol). Note that the data given has been perturbed, so looking up the answer is probably not a good idea. AHsublimation 61 kJ/mol Ionization energy = 383 kJ/mol Rb (s) Rb (g) H - H (g) Bond energy = 416 kJ/mol H (g) Electron affinity = -93 kJ/mol RbH (s) AHOf = -72 kJ/mol kJ/mol

Please provide detailed solution and give the explanation of the concept...

PCl3(g) + Cl2 (g) --><-- PCl5(g) A. Use a bond energy to find deltaHo for the reaction. Cl - Cl, 339 kJ/mole; P - Cl, 289 kJ/mole. B. Calculate the value of deltaSo for the reaction.                                                  So(PCl5) = 364.2 J/mol * K; So(PCl3) = 311.6 J/mol * K; So(Cl2) = 165.2    J/mol * K.

Determine the enthalpy of reaction (in kJ) for the following equation using bond enthalpies. N2 (g) + O2 (g) → 2 NO2- (aq)

Calculate the standard energy of formation of CI0 from the following bond energies. f Cl2: 242.7 kJ/mol O2: 498.7 kJ/mol C10: 206 kJ/mol kJ/mol

9.26 Calculate the lattice energy of calcium chloride given that the heat of sublimation of Ca is 121 kJ/mol and AHf°(CaCl₂)= -795 kJ/mol. (See Tables 8.2 and 8.3 for other data.)

Which of the following ionic compounds has the largest lattice formation enthalpy (lattice energy); that is, which compound is most favorable to a stable lattice? A) Csl B) Lil C) LiF D) CsF E) MgO

Calculate the lattice energy of KCI(s) using the following thermodynamic data (all data is in kJ/mol) K(s) AHsublimation = 69 kJ/mol K(g) Ionization energy = 399 kJ/mol CI-CI(g) Bond energy = 223 kJ/mol Cl(g) Electron affinity = -369 kJ/mol KCI(s) AH°r= -457 kJ/mol kJ/mol

Use average bond enthalpy values to calculate ΔrH for the following reaction: 2NH3 + 3Cl2 → N2 + 6HCl