Calculate the lattice energy for LiBr(s) given the following: sublimation energy for Li(s) +166 kJ/mol ΔHf for Br(g) +97 kJ/mol first ionization energy of Li(g) +520. kJ/mol electron affinity of Br(g) –325 kJ/mol enthalpy of formation of LiBr(s) –351 kJ/mol

Given the following enthalpy values for Na and Cl, calculate the lattice energy of NaCl using the Born- Haber cycle. for Na = 107.5 kJ/mole for Na = 495.8 kJ/mole for Cl, = 242.6 kJ/mole AHO sublimation AH ionization energy AHO dissociation energy ΔΗΡ electron affinity TOr Cl, = -348.6 kJ/mole AH formation for NaCl(s) = -4l1.2 kJ/mole

Lattice energies and electron affinities are very difficult to measure experimentally. Fairly reliable theoretical values for lattice energies can be calculated from simple models. The Born-Haber cycle can then be used to estimate electron affinities. Use the data below to calculate the electron affinity of iodine (in kJ mol-1), where a negative EA value indicates a favourable anion formation. Name of enthalpy change Value in kJ/mol Formation enthalpy of Cal2(s) -533.46 Lattice energy of Cal2 -2074 Sublimation enthalpy of Ca 177.7 First ionization energy of Ca 589.8 Second ionization energy of Ca 1145.4 Sublimation enthalpy of I2 62.44 Bond energy of l2 151.1

The enthalpy of the formation of ammonium chloride is AH° = -175.9 kJ/mol at 25°C. Calculate the change in energy (AE) for this reaction. NH3(g). + HCl(g) → NH,Cl(s)

Consider the following information. • The lattice energy of CsCl is AHLattice = -657 kJ/mol. • The enthalpy of sublimation of Cs is AHsub = 76.5 kJ/mol. The first ionization energy of Cs is IE₁ = 376 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, AH, for CsCl(s). AH₁ = kJ/mol

Place the following elements in... increasing electronegativity:    Se  Cl  S increasing ionization energy:  Al  P  Sr increasing affinity for electrons: K  Ge  Cl increasing lattice energy: Na2S  KCl  MgS

Given the quantities below, what is the lattice energy of AgCl? Select the correct answer below: -680 kJ/mol -917 kJ/mol -895 kJ/mol O-1742 kJ/mol Ionization energy of Ag = 734 kJ/mol Sublimation energy of Ag = 287 kJ/mol Dissociation energy of Cl₂ = 242 kJ/mol Enthalpy of formation of AgCl = -127 kJ/mol Electron affinity of Cl = -352 kJ/mol

1) Calculate the lattice energy for NaCl(s) using a Born-Haber cycle and the following information: NaCl(s) → Nat(g) + Cl-(g) Na(s) + 1/2 C12(g) → NaCl(s) Na(s) → Na(g) Na(g) → Na+(g) + e- 1/2 C12(g) → Cl(g) Cl(g) + e- → Cl-(g) ? -411.0 kJ/mol +107.3 kJ/mol +495.8 kJ/mol +121.7 kJ/mol -348.6 kJ/mol

Question 5

Lattice energies and electron affinities are very difficult to measure experimentally. Fairly reliable theoretical values for lattice energies can be calculated from simple models. The Born-Haber cycle can then be used to estimate electron affinities. Use the data below to calculate the electron affinity of iodine (in kJ mol-1), where a negative EA value indicates a favourable anion formation. Name of enthalpy change Value in kJ/mol Formation enthalpy of Cal 2(s) -533.46 Lattice energy of Cal2 -2074 Sublimation enthalpy of Ca 177.7 First ionization energy of Ca 589.8 Second ionization energy of Ca 1145.4 Sublimation enthalpy of I2 62.44 Bond energy of 12 151.1 -293 +265 -599 -482