![]() ![]() The rates at which the two lattice energies fall as you go down the Group depends on the percentage change as you go from one compound to the next. The inter-ionic distances in the two cases we are talking about would increase from 0.365 nm to 0.399 nm - an increase of only about 9%. For the sake of argument, suppose that the carbonate ion radius was 0.3 nm. I can't find a value for the radius of a carbonate ion, and so can't use real figures. Although the inter-ionic distance will increase by the same amount as you go from magnesium carbonate to calcium carbonate, as a percentage of the total distance the increase will be much less. In the carbonates, the inter-ionic distance is dominated by the much larger carbonate ion. In the oxides, when you go from magnesium oxide to calcium oxide, for example, the inter-ionic distance increases from 0.205 nm (0.140 + 0.065) to 0.239 nm (0.140 + 0.099) - an increase of about 17%. The oxide ion is relatively small for a negative ion (0.140 nm), whereas the carbonate ion is large (no figure available). The lattice enthalpies fall at different rates because of the different sizes of the two negative ions - oxide and carbonate. The inter-ionic distances are increasing and so the attractions become weaker. The lattice enthalpies of both carbonates and oxides fall as you go down the Group because the positive ions are getting bigger. If the attractions are large, then a lot of energy will have to be used to separate the ions - the lattice enthalpy will be large. Forces of attraction are greatest if the distances between the ions are small. The size of the lattice enthalpy is governed by several factors, one of which is the distance between the centres of the positive and negative ions in the lattice. If you think carefully about what happens to the value of the overall enthalpy change of the decomposition reaction, you will see that it gradually becomes more positive as you go down the Group.Įxplaining the relative falls in lattice enthalpy The oxide lattice enthalpy falls faster than the carbonate one. You can apply Hess's Law to this, and find two routes which will have an equal enthalpy change because they start and end in the same places.įor reasons we will look at shortly, the lattice enthalpies of both the oxides and carbonates fall as you go down the Group. The cycle we are interested in looks like this: The term we are using here should more accurately be called the "lattice dissociation enthalpy". ![]() In that case, the lattice enthalpy for magnesium oxide would be -3889 kJ mol -1. Lattice enthalpy is more usually defined as the heat evolved when 1 mole of crystal is formed from its gaseous ions. Here's where things start to get difficult! If you aren't familiar with Hess's Law cycles (or with Born-Haber cycles) and with lattice enthalpies (lattice energies), you aren't going to understand the next bit. ![]()
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |