5. Consider the following information: 1st ionization energy of Na(g) = 495.8 kJ/mol Bond dissociation energy of O2(g) = 498.4 kJ/mol 1st electron affinity of O(g)=-142.5 kJ/mol 2nd electron affinity of O¹(g) = 844 kJ/mol Lattice energy of Na2O(s) = -2608 kJ/mol Enthalpy of formation of Na2O(s) = -416 kJ/mol a Draw the Born-Haber cycle for Na₂O(s). b Calculate the unknown. 120 C Draw the Lewis symbol for Na₂O.

Use the data provided below to calculate the lattice energy of RbCl. Is this value greater or less than thelattice energy of NaCl? Explain.Electron affinity of Cl = –349 kJ/mol1st ionization energy of Rb = 403 kJ/molBond energy of Cl2 = 242 kJ/molSublimation energy of Rb = 86.5 kJ/molΔHf [RbCl (s)] = –430.5 kJ/mol

Given the following information, construct a Born-Haber cycle to calculate the lattice energy of CrCl₂I(s): Net energy change for the formation of CrCl₂I(s) = -420 kJ/mol Bond dissociation energy for I2(g) = +243 kJ/mol Bond dissociation energy for Cl2(g) for Cl2(g) for 12(g) = +151 kJ/mol Heat of sublimation for I2(s) = +62 kJ/mol Heat of sublimation for Cr(s) = +397 kJ/mol = E₁₁ for Cr(g) = 652 kJ/mol E₁₂ for Cr(g) == 1588 kJ/mol E₁3 for Cr(g) = 2882 kJ/mol Eea for Cl(g)=-349 kJ/mol === Eea for I(g) = -295 kJ/mol

Consider a hypothetical ionic compound AB (comprised of A* and B ions). Given the following enthalpy data and using a Born-Haber cycle calculation, predict AHiattice in kJ mol. AG) + Bs) - AB 4H = -384 kJ mol As) → Ag) ArH = 105 kJ mol B(s) - Bg) A,H = 101 kJ mol1 First ionization energy of Ag) = 501 kJ mol" Answes=? Electron affinity enthalpy of Bro (exothermic) = -348 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

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

8. Given the following information: Li(s) HI(g) → H(g) + I(g) enthalpy of sublimation of Li(s) = 166 kJ/mol bond energy of HI = 295 kJ/mol Li(g) Li(g) → Li"(g) + e ionization energy of Li(g)= 520. kJ/mol I(g) + e — Г(g) electron affinity of I(g) = -295 kJ/mol Li"(g) + I(g) → LiI(s) lattice energy of LiI(s) = -737 kJ/mol H2(g) → 2H(g) Calculate the change in enthalpy for: bond energy of H2 = 432 kJ/mol 2Li(s) + 2HI(g) –→ H2(g) + 2LİI(s) a. 330 kJ b. –534 kJ c. -483 kJ d. -984 kJ e. none of these

Calculate the lattice energy of NaBr(s), given the following thermochemical equations, where A/E and AEA are ionization energy and electron affinity, respectively. Na(s)Na(g) AH = +107 kJ Na(g) Nat(g) + e A/E = +496 kJ -> 1/2 Br₂(g) → Br(g) AHf = +112 kJ - Br(g) + e¯ → Br¯(g) AEA = -325 kJ Na(s) + 1/2 Br₂(g) → NaBr(s) AH = -361 kJ ->> - -1401 kJ -751 kJ +29 kJ -29 kJ +751 kJ

g 44.956 mol atomic mass electronegativity 1.36 kJ 18.1 mol electron affinity kJ 633.1 mol ionization energy kJ 16. mol heat of fusion Does the following reaction absorb or release energy? O release O absorb (1) Sc(g) + e Sc (g) O Can't be decided with the data given. Is it possible to calculate the amount of energy absorbed or released by reaction (1) using only the data above? O yes O no If you answered yes to the previous question, enter the amount of energy absorbed or released by reaction (1): O kJ/mol Does the following reaction absorb or release energy? O release O absorb (2) Sc (g) - Sc (g) + e O Can't be decided with the data given. Is it possible to calculate the amount of energy absorbed or released by reaction (2) using only the data above? O yes O no If you answered yes to the previous question, enter the amount of energy absorbed or released by reaction (2): O kJ/mol

Given the following information: Li(s) → Li(g) Li(s) = 161 kJ/mol Heat of sublimation of HCI(g) → H(g) + Cl(g) Bond energy of HCI = 427 kJ/mol Li(g) → Li+(g) + e- lonization energy of Li(g) = 520. kJ/mol CI(g) + e- → CI-(g) Electron affinity of Cl(g) = -349 kJ/mol Li+(g) + Cl-(g) →→ LiC(s) Lattice energy of LİCI(s) = -829 kJ/mol H2(g) → 2H(g) Bond energy of H2 = 432 kJ/mol calculate the net change in energy for the reaction 2Li(s) + 2HCI(g) → 2LİCI(s) + H2(g) -179 kJ 362 kJ -70 kJ -572 kJ None ese