Question

What is the lattice energy of CaF2 in kilojoules per mole given that the distance between the calcium ion and fluoride ion is 2.01 Å and the charges on the ions are +2 and -1 respectively?

Answer 1

To calculate the lattice energy of CaF, we can use the Born-Haber cycle or the Coulomb's Law equation for lattice energy:Lattice Energy =  k * Q * Q  / rwhere:k = Coulomb's constant  8.9875517923  10 Nm/C Q = charge on the calcium ion  +2 Q = charge on the fluoride ion  -1 r = distance between the ions  2.01 , which needs to be converted to meters First, let's convert the distance between the ions from  to meters:1  = 1  10 meters2.01  = 2.01  10 metersNow, we can plug the values into the equation:Lattice Energy =  8.9875517923  10 Nm/C  *  +2  *  -1  /  2.01  10 m Lattice Energy = -8.9875517923  10 Nm/C * 2 / 2.01  10 mLattice Energy = -17.9751035846  10 Nm/C / 2.01  10 mLattice Energy = -8.947  10 JNow, we need to convert the energy from Joules to kilojoules per mole:1 J = 0.001 kJ1 mole = 6.022  10 particlesLattice Energy = -8.947  10 kJ/particle * 6.022  10 particles/moleLattice Energy = -5386.4 kJ/molTherefore, the lattice energy of CaF is approximately -5386.4 kJ/mol.