Born haber cycle of CaCal2
Answers
Answer:
The Born Haber cycle for the formation of calcium chloride (CaCl2) form its constituent elements involves following steps:¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤¤
Answer:
The Born Haber cycle for the formation of calcium chloride (CaCl2) form its constituent elements involves following steps:
Atomization enthalpy of Calcium: This step involves the conversion of solid calcium to gaseous state.
Ca(s) = Ca(g), ΔHa°(Ca) = 178 kJ/mol
Ionization enthalpy of Calcium: Calcium forms di-positive ion (Ca2+), therefore the energy involve in the removal of first electron is called as first ionization energy and valued 590 kJ/mol. The removal of second electron requires more energy as it’s difficult to remove electron from a cation. Hence the second ionization energy for calcium ion is 1145 kJ/mol.
Ca(g) = Ca+(g) + e-, ΔHIE°= 590 kJ/mol
Ca+(g) = Ca2+(g) + e-, ΔHIE°= 1145 kJ/mol
Atomization enthalpy of Chlorine: This step involves dissociation of Cl2(g) into Cl(g) atoms. The reaction enthalpy is half of the bond dissociation enthalpy of chlorine.
½ Cl2(g) = Cl(g), ΔHa°= ½ΔHCl-Cl°= 121 kJ/mol
For the formation of calcium chloride, two Cl(g) requires, therefore total atomization enthalpy will be double that is 242 kJ/mol.
Electron affinity of Chlorine: This is the amount of energy released during the addition of electron in an isolated neutral gaseous chlorine atom.
Cl(g) + e- = Cl-(g), ΔHEA°= -364 kJ/mol
Lattice enthalpy: The combination of one Ca2+ ion and two chloride ions (Cl-)to form one mole of calcium chloride release lattice energy (ΔHLE).
Ca2+ (g) + 2Cl-(g) = CaCl2(s), ΔHLE°= -364 kJ/mol
Ca(s) + Cl2(g) = CaCl2(s), ΔHf°
The heat of formation of calcium chloride can be calculated by using the Born Haber cycle in which the sum of enthalpy in a cycle is zero. By applying the Hess’s Law,
Heat of formation (ΔHfo) = Heat of atomization (ΔHa°) + Dissociation energy (ΔHd°) + (sum of Ionization energies) + (sum of Electron affinities) + Lattice energy (ΔHLE°)
ΔHfo (CaCl2) = 178 kJ/mol + 2*121 kJ/mol + 2*(-364 kJ/mol) + 590 kJ/mol + 1145 kJ/mol + (-2223 kJ/mol) = 796 kJ / mol.
ΔHfo (CaCl2) = 796 kJ / mol.