Calculate the lattice enthalpy of silver chloride from the following data.
Enthalpy of formation of Ag(g) : +284 kJ.mol^-¹
First ionization energy of Ag(g) : +731 kJ.mol^-¹
Enthalpy of formation of Cl(g) : +122 kJ.mol^-¹
Electron affinity of Cl(g) : +349 ( $\Delta$ H = -349)
kJ.mol^-¹
Enthalpy of formation of AgCl(s) : -127 kJ.mol^-¹

Question

Calculate the lattice enthalpy of silver chloride from the following data.
Enthalpy of formation of Ag(g) : +284 kJ.mol^-¹
First ionization energy of Ag(g) : +731 kJ.mol^-¹
Enthalpy of formation of Cl(g) : +122 kJ.mol^-¹
Electron affinity of Cl(g) : +349 (

\Delta

H = -349)
kJ.mol^-¹
Enthalpy of formation of AgCl(s) : -127 kJ.mol^-¹

Quizplus · Accepted Answer

The lattice enthalpy of silver chloride can be calculated using the Born-Haber cycle, which involves the following steps:1. Sublimation of Ag(s) to Ag(g): +284 kJ/mol2. Ionization of Ag(g) to Ag+(g): +731 kJ/mol3. Formation of Cl(g) from Cl2(g): +122 kJ/2 = +61 kJ/mol (since the given value is for diatomic chlorine, and we need the value for a single atom)4. Addition of an electron to Cl(g) to form Cl-(g): -349 kJ/mol5. Formation of AgCl(s) from its elements: -127 kJ/molThe lattice enthalpy is the sum of these steps, but with the formation enthalpy of AgCl(s) subtracted from the sum of the other steps, as it represents the overall process of forming AgCl(s) from its elements in their standard states.Thus, the calculation is: 284 kJ/mol + 731 kJ/mol + 61 kJ/mol - 349 kJ/mol - (-127 kJ/mol) = 915 kJ/mol.

Calculate the Lattice Enthalpy of Silver Chloride from the Following Δ\DeltaΔ

Calculate the Lattice Enthalpy of Silver Chloride from the Following $\Delta$