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Forming Mg²⁺ and O²⁻ seems costly. In kJ/mol: sublimation of magnesium +148; half the bond enthalpy…
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Forming and seems costly. In kJ/mol: sublimation of magnesium ; half the bond enthalpy of oxygen ; first and second ionisation of magnesium and ; first and second electron gain of oxygen and . The lattice enthalpy of MgO is . Find the energy needed to make the gaseous ions from the elements, and the enthalpy of formation of MgO.
Need a reminder? The key ideaShow
Born-Haber cycle: enthalpy of formation sublimation half the bond enthalpy ionisation enthalpy electron gain enthalpy lattice enthalpy. The lattice enthalpy is the energy needed to separate one mole of an ionic solid into its gaseous ions. It grows with the charges of the ions and as the ions get smaller.
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Ionisation. kJ/mol.
Step 2. Electron gain. kJ/mol. Adding the second electron to is uphill.
Step 3. The gaseous ions. kJ/mol.
Step 4. Formation. kJ/mol.
Step 5. Why. The doubly charged ions cost a lot, but they attract four times as strongly. The large lattice enthalpy more than pays.
Answer: kJ/mol to make the ions; kJ/mol for MgO.
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