Calculating Lattice Energy Using the Born Haber Cycle

University:
University of Arkansas
Course:
CHEM 1203 | Chemistry for Majors I
Academic year:

2019
Views:

195

Pages:

1
Author:

BobN

In this example problem, we will calculate the lattice energy for an ionic compound using the Born-Haber cycle. Our problem reads: Calculate the lattice energy for lithium oxide, given the sublimation energy for lithium, the first ionization energy of lithium, the bond energy for an oxygen-oxygen double bond, the first electron affinity for oxygen, the second electron affinity for oxygen, and the enthalpy of formation for lithium oxide. To solve this problem, we're going to add together multiple reactions to produce our enthalpy of formation equation, which is: 2 lithium solid + 1/2 oxygen gas → lithium oxide solid First, we'll use the sublimation energy of lithium to go from lithium solid to lithium gas. In our enthalpy of formation equation, we need two moles of lithium, so we'll multiply this equation by two, which also means that we need to multiply the sublimation energy by two. Next, we're going to remove an electron from the gaseous lithium atom. This corresponds to the first ionization energy of lithium. We'll need to multiply this equation by two again, which means we also need to multiply the first ionization energy of lithium by two. The next step is to break the oxygen-oxygen double bond in oxygen gas, so that we have two gaseous oxygen atoms in our equation. We only need 1/2 mole of oxygen gas, so we'll multiply this reaction by 1/2, which means we also multiply the bond energy of the oxygen-oxygen double bond by 1/2. Next, we need to add an electron to our gaseous oxygen atom. This corresponds to the electron affinity of oxygen. Our first electron affinity is negative, which means this is an exothermic process. But our second electron affinity is positive, so it requires energy to add the second electron to the gaseous oxygen anion to produce O2- gas. Now, we have two gaseous lithium cations and a gaseous oxygen anion, and these gaseous cations and anions will combine together to produce our lithium oxide solid. The energy released during this process is our lattice energy. When we add together all of these reactions, we're going to get the reaction for the enthalpy of formation for lithium oxide. So, according to Hess's law, if we add together the enthalpy changes for all of these reactions, we should get the enthalpy of formation for lithium oxide. We can solve this equation for the lattice energy of lithium oxide, and we should get -2859 kilojoules.

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