THERMAL STABILITY OF GROUP 2 CARBONATES AND NITRATES
This page discusses the effect of heat on carbonates and nitrates of Group 2 elements: beryllium, magnesium, calcium, strontium, and barium. Describe and explain how the thermal stability of compounds changes as you move down the group.
The effect of heat on group 2 carbonates
All carbonates in this group thermally decompose to metal oxide and carbon dioxide gas. Thermal decomposition is the term for the cleavage of a compound when heated.
All these carbonates are white solids and the oxides produced are also white solids.
If "X" represents one of the elements:
Moving down the group, the carbonates must be heated more before they decompose.
The effect of heat on group 2 nitrates
All nitrates in this group thermally decompose to metal oxide, nitrogen dioxide and oxygen.
Nitrates are white solids and the oxides produced are also white solids. The brown gas nitrogen dioxide is emitted along with the oxygen. Magnesium and calcium nitrates usually have water of hydration, and the solid may dissolve in its own water of hydration to form a colorless solution before starting to decompose.
Again, if "X" represents one of the elements:
2X (NOT3)2(S)2XO(s) + 4NO2(g) +O2(Grama)
As the group shuts down, nitrates also need more heating before they break down.
Both carbonates and nitrates become more thermally stable further down the group. The bottom ones need to get hotter than the top ones before they break.
This page offers two different perspectives on the problem. You need to figure out what your examiners are likely to expect from you, so you don't get into more difficult stuff than you really need to. You should look at your resume and past exams.along with your branded systems
|To use:If you are working towards a UK-based exam (A-level or equivalent) and you don't have copies of itResume and previous workFollow this link to find out how to get them.|
Detailed explanations are given for carbonates because diagrams are easier to draw and their equations are also easier. Exactly the same arguments apply to nitrates.
Explanation of the trend in the polarization ability of positive ions
A small 2+ ion packs a lot of charge into a small volume of space. has a highcharge densityand has a distinct distorting effect on any nearby negative ions.
A larger 2+ ion has the same charge spread over a larger volume of space. Its charge density is lower and causes less distortion to nearby negative ions.
The structure of the carbonate ion.
If you were to solve the structure of a carbonate using "dots and crosses" or a similar method, you would probably get the following:
This shows two single and one double carbon-oxygen bonds, with two of the oxygen atoms each carrying a negative charge. Unfortunately, in true carbonate ions all the bonds are identical and the charges are distributed throughout the ion even though they are concentrated on the oxygen atoms. We say that the rates arerelocated.
This is a slightly more complicated version of the bond you might have found in benzene or ions like ethanoate. For the purposes of this topic, you don't need to understand how this link came to be.
|To use:If you are interested, you can follow these linksBenzenethe stopOrganic acids. Any of these links will likely take you on a rather lengthy detour!|
The diagram below shows the delocalized electrons. The shading is to show that they are more likely to be found near oxygen atoms than near carbon.
Polarization of Carbonations
Now imagine what happens when this ion is placed next to a positive ion. The positive ion attracts the delocalized electrons on the carbonate ion. The carbonate ion is polarized.
When heated, carbon dioxide is released to leave the metal oxide.
How much you need to heat the carbonate before this happens depends on how polarized the ion was. If it is heavily dyed, it needs less heat than if it is lightly dyed.
The smaller the positive ion, the greater the charge density and the greater the effect it has on the carbonate ion. As positive ions increase in size further down the group, they have less of an impact on nearby carbonate ions. To compensate for this, you need to heat the compost more to release the carbon dioxide and leave the metal oxide behind.
In other words, as you move down the group, the carbonates become more thermally stable.
What about nitrates?
The reasoning here is exactly the same. The small positive ions at the top of the group polarize the nitrate ions more than the larger positive ions at the bottom. Drawing diagrams to show that this is happening is much more difficult because the process has interactions involving more than one nitrate ion. You are not expected to try to draw this on a test.
Explain the trend in terms of process energy.
Observing enthalpy changes
If you calculate the enthalpy changes for the decomposition of the various carbonates, you will find that all the changes are fairly endothermic. This implies that the reactions probably need constant heating for them to occur.
|To use:If you are not satisfied with the enthalpy changes, you can examine thempower sectionda Chemguide, amy chemistry book.|
Enthalpy changes (in kJ mol-1) which I calculated from the enthalpy changes of formation given in the table. Figures to calculate the beryllium carbonate value were not available. Remember that the reaction we are talking about is this:
You can see that the reactions become more endothermic as you progress through the group. This is exactly what one would expect as carbonates become more thermally stable. You have to add more and more heat energy to break it down.
Explanation of enthalpy changes
This is where it starts to get tricky! If you are not familiar with Hess law cycles (or Born-Haber cycles) and lattice enthalpies (lattice energies), you will not understand the following. Don't waste time looking at it.
Using an enthalpy cycle
You can investigate to find the underlying causes of the increasingly endothermic changes as you move up the group by tracing an enthalpy cycle that includes the lattice enthalpies of metal carbonates and metal oxides.
Confusingly, there are two ways to define lattice enthalpy. To simplify the argument mathematically, I'll use the less common version (as far as UK A Level study programs are concerned) for the rest of this page:
network enthalpyis the heat required to split a mole of crystal in its standard state into its individual gaseous ions. For example, for magnesium oxide, the heat required to make 1 mole of this change:
MgO(s)Magnesium2+(g) +O2-(g) LE= +3889 kJ mol-1
|To use:Lattice enthalpy is most generally defined as the heat generated when 1 mol of crystal is formed from its gaseous ions. In this case, the lattice enthalpy for magnesium oxide would be -3889 kJ mol-1. The term we use here should be more precisely "enthalpy of network dissociation".|
The loop we are interested in looks like this:
You can apply Hess' law to this and find two routes that have the same enthalpy change because they start and end at the same places.
For reasons we'll see shortly, the lattice enthalpies of both oxides and carbonates decrease as you move down the group. But they don't fall in the same proportion.
The enthalpy of the oxide lattice drops faster than that of the carbonate. If you think carefully about what happens to the total value of the enthalpy change of the decomposition reaction, you will find that it gradually becomes more positive as you move down the group.
Explanation of the relative drop in lattice enthalpy
The magnitude of lattice enthalpy is determined by several factors, one of which is the distance between the centers of positive and negative ions in the lattice. Attractive forces are greatest when the distances between ions are small. If the attractions are large, a lot of energy must be expended to separate the ions: the lattice enthalpy will be large.
The lattice enthalpies of carbonates and oxides decrease as you go down the group because the positive ions get bigger and bigger. Interionic distances increase and attractions decrease.
The lattice enthalpies fall at different rates due to the different sizes of the two negative ions, oxide and carbonate. The oxide ion is relatively small (0.140 nm) for a negative ion, while the carbonate ion is large (numbers not available).
In oxides, for example, going from magnesium oxide to calcium oxide, the interion distance increases from 0.205 nm (0.140 + 0.065) to 0.239 nm (0.140 + 0.099), an increase of about 17%.
In carbonates, the interion distance is dominated by the much larger carbonate ion. Although the distance between the ions increases as you move from magnesium carbonate to calcium carbonate,as a percentage of the total distancethe increase will be much smaller.
Some invented characters show this clearly.
I can't find a value for the radius of a carbonate, so I can't use real numbers. For the sake of argument, assume that the radius of the carbonate ion is 0.3 nm. The interion 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%.
The rates at which the two grid energies decay as you move down the group depends on the percentage change as you move from one connection to the next. Based on this, the enthalpies of the oxide lattice should drop faster than those of the carbonates.
What about nitrates?
The nitrate ion is larger than an oxide ion, so its radius tends to dominate the interionic distance. The lattice enthalpy of the oxide will again fall faster than that of the nitrate. If you were to create a loop further down the page, the same arguments would apply.
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Questions to test your understanding
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Questions about the thermal stability of group 2 carbonates and nitrates
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© Jim Clark 2002 (last modified December 2021)
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