The thermal stability of nitrates and carbonates (2023)

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This page examines the effect of heat on carbonates and nitrates of Group 2 elements (beryllium, magnesium, calcium, strontium and barium). Explain how the thermal stability of compounds changes in the group.

The effect of heat on group 2 carbonates

All carbonates in this group thermally decompose to metal oxide and carbon dioxide gas. The term "thermal decomposition" describes the breakdown of a compound by heating.

All Group 2 carbonates and their resulting oxides exist as white solids. When "X" represents one of the elements, this decomposition is described as follows:

\[XCO_3(s) \rightarrow XO(s) + CO_2(g)\]

In the lower group, carbonates require more heat to decompose.

Carbonates become more thermally stable in the group.

The effect of heat on group 2 nitrates

Group 2 nitrates thermally decompose to metal oxide, nitrogen dioxide and oxygen gas. These compounds are white solids and brown nitrogen dioxide, and also emit oxygen gases when heated. Magnesium and calcium nitrates normally crystallize with water, and the solid is allowed to dissolve in its own water of crystallization to form a colorless solution before it begins to decompose.

Again, if "X" represents one of the elements:

\[ 2X(NO_3)_2(s) \rightarrow 2XO(s) + 4NO_2(g) + O_2 (g)\]

Below the group, nitrates also require more heating before they decompose.

Group 2 nitrates also become more thermally stable in the group.

Summary

Both the carbonates and nitrates of the Group 2 elements become more thermally stable in the group. Larger compounds found below require more heat than lighter compounds to break down.

Explanations

The structure of the carbonate ion.

An abbreviated structure for the carbonate ion is given below:

This structure has two carbon-oxygen single bonds and one double bond, with two of the oxygen atoms each carrying a negative charge. In true carbonate ions, all the bonds are identical and the charges are distributed throughout the ion, with the greatest density concentrated on the oxygen atoms. In other words, loads are delocalized.

The diagram below shows the delocalized electrons. The shading is intended to show that there is more electron density around oxygen atoms than around carbon.

Polarization of Karbonations

When this ion is placed next to a cation, such as a group 2 ion, the cation attracts the delocalized electrons in the carbonate ion, pulling the electron density toward it. The carbonate ion is polarized.

When the carbonate is heated, carbon dioxide is released leaving behind the metal oxide.

The amount of heating required depends on the degree of polarization of the ion. More polarization requires less heat. The smaller the positive ion, the greater the charge density and the greater the effect it has on the carbonate ion. As the positive ions in the group increase in size, they affect the carbonate ions next to them less. In order for the carbon dioxide to leave the metal oxide, more heat must be supplied.

In other words, carbonates become more thermally stable in the group.

What about nitrates?

The argument is exactly the same for the group 2 nitrates. The small cations at the top of the group polarize the nitrate ions more than the larger cations at the bottom. This process is much more difficult to visualize due to the interactions of multiple nitrate ions.

enthalpy changes

The enthalpy changes for the decomposition of the various carbonates show that the reactions are strongly endothermic, implying that the reactions likely require constant heating to proceed. Remember that the reaction in question is as follows:

\[XCO_{3(s)} \rightarrow XO_{(s)} + CO_{2(g)}\]

The calculated enthalpy changes (in kJ mol^-1) are given in the table below (no data available for beryllium carbonate).

Carbonate	\(\DeltaH\)
MgCO₃	+117
CaCO₃	+178
SrCO₃	+235
Base₃	+267

The reactions are expected to be more endothermic as carbonates become more thermally stable, as discussed above.

Explanation of the enthalpy changes

This is where it starts to get tricky! If you are not familiar with Hessian 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 plotting an enthalpy cycle that includes the lattice enthalpies of metal carbonates and metal oxides.network enthalpyis the heat required to split one mole of crystal in its standard state into its individual gaseous ions. For example, with magnesium oxide, it is the heat required to make 1 mole of this change:

\[ MgO_{(s)} \rightarrow Mg^{2+}_{(g)} + O^{2-}_{(g)}\]

Net energy (LE): + 3889 kJ/mol

Lattice enthalpy is generally defined as the heat generated when 1 mole 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 "network decoupling enthalpy".

The loop we are interested in looks like this:

You can apply Hess's 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 in a moment, the lattice enthalpies of both oxides and carbonates decrease as you go down the group. But they don't fall in equal proportion. The enthalpy of the oxide lattice decreases faster than that of the carbonate. If you think carefully about what happens to the total enthalpy change value of the decomposition reaction, you'll find that it gradually becomes more positive as you move down the group.

Explanation of the relative drop in lattice enthalpy

The enthalpy size of the lattice is determined by several factors, one of which is the distance between the centers of the positive and negative ions in the lattice. Attraction forces are greatest when the distances between the 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 larger and larger. Interionic distances increase and attractions decrease.

	Radioion (nm)
Magnesium²⁺	0,065
California²⁺	0,099
Ö^2-	0,140
CO₃²^-	?

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 for a negative ion (0.140 nm) while the carbonate ion is large (numbers not available).

For oxides, for example, when changing from magnesium oxide to calcium oxide, the distance between the ions 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 interionic distance is dominated by the much larger carbonate ion. Although internal distance increases at the same rate as you move from magnesium carbonate to calcium carbonate, the increase as a percentage of the total distance will be much less.

Some fictional 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 distances between them 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 web energies fall as you move down the group depend on the percentage change as you move from one link to the next. As a result, the enthalpies of the oxide lattice tend to fall 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 distance between the ions. Again, the lattice enthalpy of the oxide will fall faster than that of the nitrate. If you were to create a loop like this further up the page, the same arguments would apply.

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