NaCl(s) ---> Na+(g) + Cl-(g)
The key to understanding how to calculate the lattice energy of an ionic compound is to think about the construction of the ionic compound from two different directions: a formation reaction and the reverse of the definition of lattice energy. For the case of NaCl we can write this as
Na+(g) + Cl-(g)------- : ! : ! : !-LE : ! : ! : ! : ! : \/ : Na(s) + 1/2Cl2(g) ------> NaCl(s) Hfor
Now since energy is a state function then the path indicated by dotted lines from the formation reactants to the gaseous ions + (-LE) has to be equal to the energy required to just go from the formation reactants to the product (Hfor). This is an application of Hess's law. So now all we need to do is to complete the path from the formation reactants to the gaseous reactants. We need to take Na(s) to Na(g) which is the heat of sublimation and we then need to take Na(g) to the sodium ion which is the first ionization potential. For chlorine, we need to take the diatomic gas to the single atom gas which would be 1/2 the heat of dissociation and then we need to take the atom gas into the anion gas which would be the first electron affinity. We use the 1/2 in front because that is the number of moles of the reactant.
Then setting the cycle going from the left up and then to NaCl(s) to the cycle going to the right would be:
Hfor = Na heat of sublimation + Na's first IP + 1/2 the chlorine heat of dissociation + the chlorine first EA - LE
Now all we have to do is look up all of these items and then solve for the LE.
Practice, practice, practice.
|Now take a practice quiz to help you understand if you understand the basic concepts.|
|You must use your real name when it asks for a name.|
|The test will only submit when you have answers all of the questions correctly.|
|If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests.|