Example 2: Calculating Lattice Energy Using the Born Haber Cycle

In this example, we calculate the lattice energy of aluminum oxide using the Born-Haber cycle. Our problem reads, "Calculate the lattice energy for aluminum oxide in kilojoules per mole." We're given the sublimation energy for aluminum, the first, second, and third ionization energies of aluminum, the bond energies for an oxygen-oxygen double bond, the first and second electron affinity of oxygen, and the enthalpy of formation for aluminum oxide. To find the lattice energy, we'll need to start by writing out the standard enthalpy of formation equation for aluminum oxide. The standard enthalpy of formation is the enthalpy change associated with the formation of one mole of aluminum oxide from its constituent elements, which are aluminum and oxygen, in their standard states. The standard state of aluminum is solid; the standard state of oxygen is a diatomic gas. When we balance this equation we need to keep one mole of aluminum oxide, so we have a fractional coefficient of oxygen gas. This is allowed because we need to maintain exactly one mole of product for the standard enthalpy of formation equation. In the Born-Haber cycle, we are applying Hess's law: we need to add together multiple reaction steps to form the standard enthalpy of formation equation. The first information we were given in the problem is the sublimation energy of aluminum. Sublimation is going directly from solid to gas. This is the only equation that will contain solid aluminum so all the solid aluminum in the overall reaction must come from this equation. We need 2 moles of solid aluminum as a reactant, so we need to multiply the sublimation equation by 2 to get 2 moles of solid aluminum as a reactant. When we multiply the coefficients by 2 for the sublimation equation, we also multiply the enthalpy of sublimation by 2. The next information we're given in the problem is the first ionization energy of aluminum. The first ionization energy is the amount of energy required to remove an electron from a gaseous aluminum atom to produce a gaseous aluminum cation with a 1+ charge. This equation doesn't have any of the reactants or products that show up in the overall equation at the top. It does, however, have aluminum gas. In the sublimation equation, we added 2 moles of aluminum gas as a product, which does not show up in the overall equation. So we need to get rid of it by adding the same moles of aluminum gas as a reactant in another equation. The first ionization energy equation shows aluminum gas as a reactant. We multiply this equation by 2 so that the 2 moles of aluminum gas as a reactant cancel out the two moles of aluminum gas is a product from the previous step. The next information we're given is the second ionization energy of aluminum. The second ionization energy is the amount of energy required to remove the second electron from an already positive gaseous aluminum cation. We can use this equation to cancel out the 2 moles of Al 1+ we added as a product in the previous equation. To do this we multiply the entire equation times 2 along with its enthalpy change. The next information we're given is the third ionization energy of aluminum, which is the energy required to remove the third electron from gaseous Al 2+. To cancel out 2 moles of gaseous Al 2+ product from the previous step, we multiply this equation by 2 as well. The next information we're given in the problem is the bond energy for an oxygen-oxygen double bond. This is the amount of energy required to break the double bond and form 2 will on bond did unbonded oxygen atoms. This will be the only equation that contains 02 gas. We want 3/2 02 gas as a reactant in the overall reaction, so we need to multiply the bond energy equation by 3/2. The next information we're given in the problem is the first electron affinity of oxygen. This is the change in energy when we add an electron to a gaseous oxygen atom to produce O 1- gas. If we look closely at the previous reaction, we've added 3 oxygen gas atoms as a product--since we have 3/2 times the coefficient of 2--and these 3 gaseous atoms do not show up in the overall reaction at the top. To cancel them out, we need to multiply the first electron affinity equation by 3. The next information we're given is the second electron affinity, which is the energy changes associated with adding an electron to an already negative gaseous oxygen anion. We can multiply this equation by 3 to cancel out the 3 gaseous oxygen anions added as a product in the previous step. If we look closely, we'll see that in the ionization energy equations we add a total of 6 electrons as products and now we've added a total of 6 electrons as reactants in the electron affinity equations. These 6 electrons now cancel each other out. Our last step is to bring the gaseous cations and anions together to produce solid aluminum oxide in its crystal lattice. The enthalpy change associated with this step is the lattice energy, which is what we're solving for. This step cancels out the Al 3+ cations and the O 2- anions that were added as products in the previous steps and gives us 1 mole of aluminum oxide from our overall equation. If we look closely, everything has canceled out except for what is in our overall reaction at the top. Adding together all of these reaction steps, we get the standard enthalpy of formation equation for aluminum oxide, which has an enthalpy change of -1677 kilojoules. Since we can add all of these steps together to get the overall reaction, according to Hess's law, if we add together the enthalpy changes for all of the reaction steps we will get the enthalpy change for the overall reaction. If we solve this equation for the lattice energy, we get -15,506 kilojoules for the equation as written, which is for the formation of 1 mole of aluminum oxide.