Sublimation energy of Ca is 121 kJ/mol. Dissociation energy of is 242.8 kJ/mol, the total ionization β Thermodynamics and Thermochemistry Chemistry Question
Question
Sublimation energy of Ca is 121 kJ/mol. Dissociation energy of $Cl_2$ is 242.8 kJ/mol, the total ionization energy of Ca(g) -> Ca2+(g) is 2422 kJ/mol and electron affinity of Cl is -355 kJ/mol. Lattice energy of $CaCl_2$ is -2430.8 kJ/mol. What is Ξ΄ H for the process Ca(s) + $Cl_2$(g) -> $CaCl_2$(s)?
Answer: A
π‘ Solution & Explanation
Born-Haber: delta_f H = sub_H(Ca) + IE_total(Ca) + BE($Cl_2$) + 2*EA(Cl) + Lattice = 121 + 2422 + 242.8 + 2*(-355) - 2430.8 = 2785.8 - 710 - 2430.8 = -355 kJ/mol.
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