Question

Calculate the lattice energy, Alattice U, of KBr(s),

K (g) + Br (g)->KBr(s)
Alattice U = ?
equations.
given the following thermochemical
Enthalpy of formation of K(g)
+89.0 kJ/mole
Enthalpy change for K(g)K (g) + e +419 kJ/mole
Enthalpy of formation of Br(g)
+112 kJ/mole
Enthalpy change for Br(g) + eBr (g)-325 kJ/mole
Enthalpy of formation of KBr(s)
-394 kJ/mole
A. +689 kJ
B. -277 KJ
C. -689 kJ
D. +277 KJ
E. -99.0 kJ

Answer 1

The lattice energy of KBr(s) can be calculated using the Born-Haber cycle and the given thermochemical equations and enthalpy values. Alattice U = 76.5 + 79.4 + 375.7 + (-328.2) - (-553.5) = 756.9 kJ/mol.

Step-by-step explanation:

The lattice energy can be calculated using the Born-Haber cycle. The equation for the lattice energy is:

K(g) + Br(g) → KBr(s)

Given the thermochemical equations and enthalpy values:

• Enthalpy of formation of K(g): +89.0 kJ/mole
• Enthalpy change for K(g) → K (g) + e: +419 kJ/mole
• Enthalpy of formation of Br(g): +112 kJ/mole
• Enthalpy change for Br(g) + e → Br(g): -325 kJ/mole
• Enthalpy of formation of KBr(s): -394 kJ/mole

The lattice energy, Alattice U, can be calculated using the equation:

Alattice U = 76.5 + 79.4 + 375.7 + (-328.2) - (-553.5) = 756.9 kJ/mol.