Building and balancing chemical equations (4) Thermal decomposition

So far in this series of posts on Building and Balancing chemical equations I have discussed neutralisation, combustion and ionic precipitation.

Today, I’d like to talk about Thermal Decomposition.

First, let’s unpack the term: thermal decomposition means breaking down a compound (decomposition) into more than one product using heat (hence thermal).

Second, yes there are patterns you can observe with different types of salts.

Let’s look and consider the ways in which some salts decompose on heating.

Nitrates  NO₃^–

It really depends here on which group of the Periodic Table the nitrate is from.

You can view a YouTube video of the thermal decomposition of copper (II) nitrate here 

Note the test for Nitrogen dioxide and its colour and the colour of the copper(II) oxide that remains.

If the nitrate is a Group I nitrate then this is how it decomposes

Metal nitrate(V)  =   Metal nitrate (III)     +    oxygen

But if the nitrate is from Group 2 then this is the pattern:

Metal nitrate(V)      =     Metal oxide    +   nitrogen(IV) oxide   +   oxygen

(As we will note in other posts lithium behaves like a Group 2 metal)

Let’s now see if we can build and balance the chemical equation for the decomposition of these nitrates

Let’s start with Group 1 nitrates (except lithium)

Metal nitrate(V)  =   Metal nitrate(III)     +    oxygen

Sodium nitrate(V)  =   Sodium nitrate(III)     +    oxygen

Adding the chemical formulae gives us

Sodium nitrate(V)  =   Sodium nitrate(III)     +    oxygen

Na NO₃                         =             NaNO₂             +        O₂

Counting the particles of each element shows the oxygen need balancing so you can either halve the oxygen molecules or double the nitrates like so:

Sodium nitrate(V)  =   Sodium nitrate(III)     +    oxygen

2Na NO₃                         =             2NaNO₂             +        O₂

or

Na NO₃                         =             Na NO₂             +        1/2O₂

Here is a video of the thermal decomposition of sodium nitrate

Now let's look at a Group 2 nitrate

Here is barium nitrate Ba(NO₃)₂  lets add the formulae to the word equation

Metal nitrate(V)      =     Metal oxide    +   nitrogen(IV) oxide   +   oxygen

Ba(NO₃)₂       =          BaO        +       NO₂       +       O₂

Looking for the discrepancies in numbers of particles left and right we see two nitrogens on the left but only one on the right so double up the nitrogens there so:

Ba(NO₃)₂       =          BaO        +       2NO₂       +       O₂

That leaves six oxygens on the left and seven on the right so halve the oxygen molecules so

Ba(NO₃)₂       =          BaO        +       2NO₂       +       1/2O₂

If you are uncomfortable with “halves”  then you can now double the whole equation so

2Ba(NO₃)₂       =          2BaO        +       4NO₂       +       O₂

You should now satisfy yourself that you can make a similar attempt at the equation for the thermal decomposition of lithium nitrate (LiNO₃) into lithium oxide, nitrogen dioxide and oxygen. 

Carbonates  CO₃ ^2—   and hydrogen carbonates HCO₃ ^—

If a metal carbonate decomposes (and not all do) then it forms the metal oxide and carbon dioxide

Metal carbonate   =  metal oxide    +    carbon dioxide

Those existing solid metal hydrogen carbonates (all from Group 1) do decompose and form the oxide, water and carbon dioxide

Metal hydrogencarbonate   =  metal oxide    +    water   +   carbon dioxide

Let’s now see if we can build and balance the chemical equation for the decomposition of Group 2 carbonates.

Let’s start with the carbonate and add the formulae:

Metal carbonate   =  metal oxide    +    carbon dioxide

CaCO₃        =        CaO        +      CO₂

And you can see this is balanced but if the carbonate is from Group 1 with a single positive charge the carbonate does not decompose (unless the temperature is well above that of a Bunsen burner: see below)

Lithium carbonate is the exception so:

Lithium carbonate   =   lithium oxide  +  carbon dioxide

Li₂CO₃          =      Li₂O      +    CO₂

Again this equation is now balanced.

Sulfates SO₄^2—

Metal sulfates, if they decompose at all (and some heat is needed to decompose them) usually produce sulphur dioxide and sulphur trioxide gases

An example might be the green crystals of hydrated iron(II) sulphate

Iron(II)sulphate   =   iron(III)oxide  +   sulphur trioxide  +  sulfur dioxide

You can watch a YouTube video of this reaction here

Now if you watch this video you’ll see vapour coming off the salt initially.

This is steam from the loss of water of crystallisation.

Hydrated iron (II) sulfate has the formula FeSO₄.7H₂O

To build the equation let’s add the formulas to the word equation (we’ll ignore the water of crystallisation)

Iron(II)sulphate   =   iron(III)oxide  +   sulphur trioxide  +  sulfur dioxide

FeSO₄.     =     Fe₂O₃     +     SO₃      +   SO₂

Looking at the iron particles shows twice as many on the right hand side so double up the left side gives

2FeSO₄.     =     Fe₂O₃     +     SO₃      +   SO₂

Counting up the other particles shows the equation is balanced.

As I say other sulfates if they decompose at all have a unique equation as for example does copper sulphate CuSO₄

Copper sulphate   =   copper (II) oxide  +   sulfur trioxide   +   sulphur dioxide

But if you want to look in-depth at this reaction then there is a very good explanation of what is happening in the Chemistry Olympiad Support Booklet published on line by the British Royal Society of Chemistry.

You can access the Chemistry Olympiad site here

You will need to satisfy yourself that you can build and balance thermal decomposition equations of salts.

You could try these and see how you got on

Potassium nitrate

Zinc carbonate

Magnesium carbonate

Calcium nitrate

Copper sulphate

In later posts, I will discuss the reasons for these patterns of thermal decomposition in terms of the polarising power of the metal cations and the entropic drive given the increased number of particles produced in these reactions and the change of state of the products.