QUESTION 13 Based off the following data, what is the lattice energy of sodium chloride? AHsub...
Consider the following information. • The lattice energy of NaCl is AHlattice = –788 kJ/mol. • The enthalpy of sublimation of Na is AHsub = 107.5 kJ/mol. • The first ionization energy of Na is IE1 = 496 kJ/mol. • The electron affinity of Cl is AHEA = -349 kJ/mol. • The bond energy of Cl, is BE = 243 kJ/mol. Determine the enthalpy of formation, AHf, for NaCl(s). AH= kJ/mol
Consider the following information. The lattice energy of NaCl is ΔH lattice=−788 kJ/mol The enthalpy of sublimation of Na is ΔHsub=107.5 kJ/mol The first ionization energy of Na is IE1=496 kJ/mol. The electron affinity of Cl is ΔHEA=−349 kJ/mol. The bond energy of Cl2 is BE=243 kJ/mol. Determine the enthalpy of formation, ΔHf, for NaCl(s). ΔHf= kJ/mol
Calculating Lattice Energy Example Problem: Calculate the lattice energy of NaCl using the following information: Nas) — NagVID AH° = 109 KJ Cl2(9) ► 2010) AH = 243 KJ Nag) → Na*(g) + e AH° = 496 KJ Cle+ e - C19) AH° = -349 KJ Nas) + y2 Cl2(g) → NaCl(s) AH° = -411 KJ
calculate the lattice energy of NaCl based on the given information : ΔH°f[NaCl(s)] = -411 kJ/mol ΔH°f [Clg] = 121.5 kJ/mol ΔH°sublimation [Na] = 109 kJ/mol IE1 (Na) = 496 kJ/mol EA1 (Cl) = -349 kJ/mol
Consider the following information. • The lattice energy of KCl is AHlattice = -701 kJ/mol. • The enthalpy of sublimation of K is AHsub = 89.0 kJ/mol. • The first ionization energy of K is IE1 = 419 kJ/mol. • The electron affinity of Cl is AHEA = -349 kJ/mol. • The bond energy of Cl, is BE = 243 kJ/mol. Determine the enthalpy of formation, AHf, for KCl(s). AHư= kJ/mol
2. Use the following data to calculate the lattice energy (U) of NaCl(s) from sodium me chlorine: Enthalpy of formation (4H) for NaCl(s) - -411 kJ/mol Enthalpy of sublimation (4Hub) of Na 107.3 kJ/mol The first ionization energy of Na (E,)-495.8 kJ/mol The bond dissociation energy (D) of Clh- 243 kJ/mol The electron affinity of Cl (Eea)- 348.6 kJ/mol.
Consider the following information. The lattice energy of LiCl is ΔH lattice = −834 kJ/mol. The enthalpy of sublimation of Li is ΔH sub = 159.3 kJ/mol. The first ionization energy of Li is IE 1 = 520 kJ/mol. The electron affinity of Cl is ΔH EA = -349 kJ/mol. The bond energy of Cl2 is BE = 243 kJ/mol. Determine the enthalpy of formation, ΔHf, for LiCl(s).
2) From the following data in the Born-Haber cycle, Na(s) → Na(g) 4C12(g) → Cl(g) AH:-108 kJ/mol AHj - 495.9 kJ/mol ara =-349 kJ/mol Na(g) → Na+(g) + e- Cl(g) + e-→ Cl-(g) Na(s) +4C12(g) → NaCl(s) AHoverall =-411 kJ/mol calculate the lattice energy of NaCI.
sed on the following energy values, determine the AH- for sodium chlo Na (s) + 12 Cl2 (g) → NaCl (s) AHsub for Na: Bond Energy of Cl2: Lattice energy of NaCl: Ionization energy for Na: Electron Affinity of Cl: Na (s) → Na (g) Cl2 (g) → 2C1 (g) NaCl (s) → Na" (g) + C (g) Na (g) → Na* (g) + le Cl (g) + le → Cl (g) u u edaeftheriodietwandeta determine whe nde to determine...
need help with (CsCl) ICO Given the following enthalpies, cal Enthalpies (kJ/mol): howing enthalpies, calculate the lattice energy of Nal and CsCls), respectively. Inal(s)] = -288; AH (Na) = 107; IE(Na) = 496: AH.(1.) = 62: D(I—I) = 149, EAT) = -295; SCI(8)] = -443; AH.(Cs) = 78: IE(Cs) = 376; D(CI_CI) = 244; EA[Cl] = -349. Na (s) + + Iz (9) Hfy Na I (3) Na(9) I (9) Nat(9) I- -288 = 1074490 +33 + 149 -295 -288x=4135...