Sol.
Sublimation Reaction of Mg :
Mg(s) -----> Mg(g) , deltaH1° = 148 KJ / mol
First ionization reaction of Mg :
Mg(g) -----> Mg+(g) + e- , deltaH2° = 738 KJ / mol
Second ionization reaction of Mg :
Mg+(g) ----> Mg2+(g) + e- , deltaH3° = 1451 KJ / mol
Bond enthalpy reaction of Cl2 :
Cl2(g) ----> 2Cl(g) , deltaH4° = 242.7 KJ / mol
Electron affinity reaction of Cl :
2Cl(g) + 2e- ----> 2Cl-(g) ,
deltaH5° = - 2 × 349 = - 698 KJ/ mol ( Here energy is released , so , sign is negative and there are 2 Cl atoms , so, energy is multiplied by 2 )
Lattice energy reaction :
Mg2+(g) + 2Cl-(g) -----> MgCl2(s) ,
deltaH6° = - 2527 KJ / mol ( Here energy is released , so, sign is negative )
Now , Enthalpy of formation reaction of MgCl2 :
Mg(s) + Cl2(g) -----> MgCl2(s)
So , deltaHf°
= deltaH1° + deltaH2° + deltaH3° + deltaH4° + deltaH5° + deltaH6°
= 148 + 738 + 1451 + 242.7 - 698 - 2527
= - 645.3 KJ / mol
M.Vammed दि me: Aolullah : Section: Roster #_to Homework 10: Due Sunday, November 17th Show your...
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
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).
3. Draw the Bom Haber Cycle and calculate the lattice energies for LiF, MgO, and CaC12 using the data provided in the table below. This can be done on a separate page if you are working off this template. AH (1/2 B.E.) Electon Affinity Ionization Sublimation AH EA Energy Ionic Compoud F 80 kJ/mol 328 kJ/mol Li 520 kJ/mol 155 kJ/mol LiF -594 kJ/mol 0 249.4 kJ/mol 141 kJ/mol (15) . Mg 738 kJ/mol (19) 148 kJ/mol MgO -601 kJ/mol...
1)a. Using the Born Haber cycle, determine the enthalpy for lattice formation of MgO. Mg (s), ΔHsub = +148 kJ/mol bond dissociation energy for O2 = +499 kJ/mol 1st ionization energy for Mg = +738 kJ/mol 1st electron affinity for O = –141 kJ/mol 2nd ionization energy for Mg = +1450 kJ/mol 2nd electron affinity for O = +844 kJ/mol MgO(s), enthalpy of formation = –602 kJ/mol 1)b. Calculate the lattice formation energy of MgO using the Madelung constant....
Using the Born-Haber cycle shown below, calculate the lattice energy for MgCl2 in kJ* mol-1 Mg**g) + 2Cl(g) 2 x-349 AH2nd le(Mg) - 1451 2xAH (01) --698 My*(g) 2013) Mg (g) + 2Cl(g) AHORE (Mg) - 738 Mg(g) + 2Cl(g) 2 x 122 2xAH CIT- +244 Mg(g) + Cl2(g) AHTE (MgCl2) Mg(s) + Clą(9) AHM9) - 148 AH, (MgCl2) --641 MgCl (s)
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Using the Born-Haber cycle shown below, calculate the lattice energy for MgCl2 in kJ* mol-1 Mg*g) + 2Cl(g) 2 -349 AH2nd (Mg) - 1451 2xAH (01) --698 Mg*9) 2C119) Mg'ig)2Cl(g) 2x+122 AH. Pot (Mg) - 738 Mg(g) + 2Cl(g) 2xAH LCI - +244 Mgig). Clz (9 Mg(s) : AH (Mg) - 148 Cl2(g) AHTE MgCI) AH. (MgCl) --641 MgCl (s)