Question

Use the following data to estimate ΔHf∘ for potassium chloride. K(s)+1/2 C₂(g)⟶KCl(s) Lattice energy −690.kJ/mol Ionization energy for K 419 kJ/mol Electron affinity of Cl −349 kJ/mol Bond energy of Cl2 239 kJ/mol Enthalpy of sublimation for K 90.kJ/mol

Answer 1

To estimate the enthalpy of formation for potassium chloride (KCl), we add together the enthalpies for sublimation of K, ionization of K, halving the dissociation of Cl2, chloride electron affinity, and the energy to form the solid lattice from gaseous ions, resulting in an estimated ΔHf° of 969.5 kJ/mol.

Step-by-step explanation:

To estimate the enthalpy of formation (ΔHf°) for potassium chloride (KCl), we can use the given thermochemical data and apply the Hess's Law and Born-Haber Cycle concepts. The formation of KCl can be analyzed by considering the steps involved and their associated energy changes:

1. Sublimation of solid potassium to gaseous potassium atoms: +90 kJ/mol.
2. Ionization of gaseous potassium atoms to potassium ions: +419 kJ/mol.
3. Dissociation of Cl2 molecule into two Cl atoms (since the given reaction involves half a molecule of Cl2, the energy required is half the bond energy): +239 kJ/mol / 2 = +119.5 kJ/mol.
4. Attachment of an electron to a Cl atom to form a chloride ion, Chloride Electron Affinity: -349 kJ/mol.
5. Formation of solid KCl from the gaseous ions, which is the opposite of the Lattice energy: +690 kJ/mol (lattice energy is given as a negative value since it is exothermic, but we need the endothermic equivalent for formation).

To calculate the ΔHf° for KCl, we sum the enthalpies of each step:

ΔHf° = 90 + 419 + 119.5 - 349 + 690 = 969.5 kJ/mol

Therefore, the estimated enthalpy of formation for potassium chloride is 969.5 kJ/mol.