Charles Collingwood narrates a short YouTube tutorial explaining the Born–Haber cycle for sodium oxide (Na2O). He outlines each enthalpy step needed to form Na2O from elemental sodium (solid) and oxygen (gas): atomization of sodium (doubled), ionization of sodium (first ionization repeated for two atoms), dissociation of O2 (½ bond enthalpy), addition of two electron affinities for oxygen, and the exothermic lattice enthalpy. He states the first ionization enthalpy of sodium as "around about 494" and summarizes that the overall formation enthalpy equals the sum of the individual steps.
Questions this video answers
Na2O.
"around about 494" (as stated verbally).
One half the bond enthalpy of O2.
Chapters
The Born Haber cycle of sodium oxide. The red line at the bottom of the cycle shows the enthalpy of formation of sodium oxide from the elements in their standard states that sodium is a solid and oxygen is a gas. The formula of sodium oxide is Na2O. And so therefore, we will need two sodium atoms for every one oxygen atom. Although here, of course, in the lattice, these are ions. It's a giant ionic lattice. So the Born harbor cycle shows us an alternative way of getting to sodium oxide. Starting with the element sodium and the element oxygen. What we need to do is to create the ions to put them in their place in the lattice. And so we follow a cycle around the outside, around the other direction. From here, starting with sodium, we turn sodium to sodium atoms, sorry, in solid form, turn to gas. And this enthalpy change is the enthalpy of atomization of sodium multiplied by two. So now we have two gaseous moles of sodium atoms. The next step is we have to ionize the sodium. To ionize the sodium, there are two sodium gaseous atoms. So we need to double the ionization enthalpy. The first ionization enthalpy of sodium, which is around about 494. The next stage is to turn the oxygen molecules into oxygen atoms. And this is done with the bond enthalpy of the of oxygen. Although bond enthalpy would give us one mole of bonds being broken, we only need half a mole of bonds to produce one mole of oxygen atoms. This is one half the bond enthalpy of oxygen. So at this stage, we have two moles of sodium ions and one mole of oxygen atoms. The next step is to turn the oxygen into oxide ions. Now, this requires the addition of two electrons. This is two electron affinities, the first electron affinity and the second electron affinity. So the next step, delta H4, is the sum of the first and second electron affinities of oxygen. This creates the ions sodium and oxide. There are two sodium ions, one oxide ion, and then we put those into the lattice. This is the exothermic lattice enthalpy of sodium oxide. Because we are going from the ions into the lattice, this releases energy. And so the enthalpy change, delta h6 in is the sum of delta h1 plus delta h2 plus delta h3 plus delta h4 plus delta h5. The Bornhaber cycle of sodium oxide.
