Question

Calculate the lattice energy of CaO given the following information: the enthalpy of formation of CaO is -635 kJ/mol, the ionization energy of Ca is 590 kJ/mol, the electron affinity of O is -141 kJ/mol, and the Ca–O bond length is 2.23 Å.

Answer 1

To calculate the lattice energy of CaO, we can use the Born-Haber cycle, which relates the lattice energy to other known thermodynamic quantities. The lattice energy  U  can be calculated using the following equation:Hf = U + IE Ca  + EA O  - E Ca2+  Ca  - E O2-  O Where:Hf = enthalpy of formation of CaO = -635 kJ/molIE Ca  = ionization energy of Ca = 590 kJ/molEA O  = electron affinity of O = -141 kJ/molE Ca2+  Ca  = energy required to remove 2 electrons from CaE O2-  O  = energy required to add 2 electrons to OFirst, we need to calculate the energy required to remove 2 electrons from Ca and the energy required to add 2 electrons to O. Since the ionization energy of Ca is given for the removal of 1 electron, we will assume that the energy required to remove 2 electrons is twice the ionization energy:E Ca2+  Ca  = 2 * IE Ca  = 2 * 590 kJ/mol = 1180 kJ/molSince the electron affinity of O is given for the addition of 1 electron, we will assume that the energy required to add 2 electrons is twice the electron affinity:E O2-  O  = 2 * EA O  = 2 *  -141 kJ/mol  = -282 kJ/molNow we can plug these values into the equation:-635 kJ/mol = U + 590 kJ/mol - 141 kJ/mol - 1180 kJ/mol + 282 kJ/molSolving for U  lattice energy :U = -635 kJ/mol - 590 kJ/mol + 141 kJ/mol + 1180 kJ/mol - 282 kJ/molU = -3424 kJ/mol + 141 kJ/mol + 1180 kJ/mol - 282 kJ/molU = -3424 kJ/mol + 1039 kJ/molU = -2385 kJ/molThe lattice energy of CaO is approximately -2385 kJ/mol.