Question

What is the lattice energy of CaCl2 given that the crystal has a distance of 2.80 Angstroms between the Ca2+ and Cl- ions and the charge on each ion is +2 and -1 respectively?

Answer 1

To calculate the lattice energy of CaCl2, we can use the Born-Lande equation:Lattice Energy  U  =  N * A *  1 - 1/n  * e^2  /  4 *  *  * r where:N = Avogadro's number  6.022 x 10^23 mol^-1 A = Madelung constant  for CaCl2, A = 1.748 n = Born exponent  for CaCl2, n  8-10, we will use n = 9 e = elementary charge  1.602 x 10^-19 C  = vacuum permittivity  8.854 x 10^-12 C / N m r = distance between the ions  2.80 Angstroms = 2.80 x 10^-10 m Now, we can plug in the values and calculate the lattice energy:U =  6.022 x 10^23 * 1.748 *  1 - 1/9  *  1.602 x 10^-19 ^2  /  4 *  * 8.854 x 10^-12 * 2.80 x 10^-10 U  2.15 x 10^-18 J/ionTo convert the lattice energy to kJ/mol, we can multiply by Avogadro's number:U = 2.15 x 10^-18 J/ion * 6.022 x 10^23 ions/molU  1294 kJ/molSo, the lattice energy of CaCl2 is approximately 1294 kJ/mol.