REACTIONS OF THE GROUP 1 ELEMENTS WITH WATER
This page looks at the reactions of the Group 1 elements - lithium, sodium, potassium, rubidium and caesium - with water. It uses these reactions to explore the trend in reactivity in Group 1.
All of these metals react vigorously or even explosively with cold water. In each case, a solution of the metal hydroxide is produced together with hydrogen gas.
This equation applies to any of these metals and water - just replace the X by the symbol you want.
In each of the following descriptions, I am assuming a very small bit of the metal is dropped into water in a fairly large container.
Details for the individual metals
Lithium's density is only about half that of water so it floats on the surface, gently fizzing and giving off hydrogen. It gradually reacts and disappears, forming a colourless solution of lithium hydroxide. The reaction generates heat too slowly and lithium's melting point is too high for it to melt (see sodium below).
Sodium also floats on the surface, but enough heat is given off to melt the sodium (sodium has a lower melting point than lithium and the reaction produces heat faster) and it melts almost at once to form a small silvery ball that dashes around the surface. A white trail of sodium hydroxide is seen in the water under the sodium, but this soon dissolves to give a colourless solution of sodium hydroxide.
The sodium moves because it is pushed around by the hydrogen which is given off during the reaction. If the sodium becomes trapped on the side of the container, the hydrogen may catch fire to burn with an orange flame. The colour is due to contamination of the normally blue hydrogen flame with sodium compounds.
Potassium behaves rather like sodium except that the reaction is faster and enough heat is given off to set light to the hydrogen. This time the normal hydrogen flame is contaminated by potassium compounds and so is coloured lilac (a faintly bluish pink).
Rubidium is denser than water and so sinks. It reacts violently and immediately, with everything spitting out of the container again. Rubidium hydroxide solution and hydrogen are formed.
Caesium explodes on contact with water, quite possibly shattering the container. Caesium hydroxide and hydrogen are formed
Summary of the trend in reactivity
The Group 1 metals become more reactive towards water as you go down the Group.
Explaining the trend in reactivity
Looking at the enthalpy changes for the reactions
The overall enthalpy changes
You might think that because the reactions get more dramatic as you go down the Group, the amount of heat given off increases as you go from lithium to caesium. Not so!
The table gives estimates of the enthalpy change for each of the elements undergoing the reaction:
Note: That's the same equation as before, but I have divided it by two to show the enthalpy change per mole of metal reacting.
You will see that there is no pattern at all in these values. They are all fairly similar and, surprisingly, lithium is the metal which releases the most heat during the reaction!
Note: Apart from the lithium value, I haven't been able to confirm these figures. For lithium, sodium and potassium, they are calculated values based on information in the Nuffield Advanced Science Book of Data (page 114 of my 1984 edition). The lithium value agrees almost exactly with a value I found during a web search. The values for rubidium and caesium are calculated indirectly from the Li, Na and K values and other information which you will find in a later table on this page.
Digging around in the enthalpy changes
When these reactions happen, the differences between them lie entirely in what is happening to the metal atoms present. In each case, you start with metal atoms in a solid and end up with metal ions in solution.
Overall, what happens to the metal is this:
You can calculate the overall enthalpy change for this process by using Hess's Law and breaking it up into several steps that we know the enthalpy changes for.
First, you would need to supply atomisation energy to give gaseous atoms of the metal.
Then ionise the metal by supplying its first ionisation energy.
And finally, you would get hydration enthalpy released when the gaseous ion comes into contact with water.
Note: There is no suggestion that the reaction actually happens by this route. All we are doing is inventing an imaginary route from the start to the end point of the reaction, and using Hess's Law to say that the overall enthalpy change will be exactly the same as we can calculate using this imaginary route. If you don't know about Hess's Law, you probably aren't likely to be making much sense of all this bit of the page anyway. If you aren't happy about enthalpy changes, you might want to explore the energetics section of Chemguide, or my chemistry calculations book.
If we put values for all these steps into a table, they look like this (all values in kJ / mol):
Note: Remember that these aren't the overall enthalpy changes for the reactions when the metal reacts with water. They are only for that part of the reaction which involves the metal. There are also changes going on with the water present - turning it into hydrogen gas and hydroxide ions. To get the total enthalpy changes, you would have to add these values in as well.
The changes due to the water will, however, be the same for each reaction - in each case about -382 kJ / mol. Adding that on to the figures in this table gives the values in the previous one to within a kJ or two. The rubidium and caesium values will agree exactly, because that's how I had to calculate them in the first table. The other three in the previous table were calculated from information from a different source.
So why isn't there any pattern in these values? If you look at the various bits of information, you will find that as you go down the Group each of them decreases:
What is happening is that the various factors are falling at different rates. That destroys any overall pattern.
It is, however, possible to look at the table again and find a pattern which is useful.
Drawing something useful from these figures
Let's take the last table and just look at the energy input terms - the two processes where you have to supply energy to make them work. In other words, we will miss out the hydration enthalpy term and just add up the other two.
Now you can see that there is a steady fall as you go down the Group. As you go from lithium to caesium, you need to put less energy into the reaction to get a positive ion formed. This energy will be recovered later on (plus quite a lot more!), but has to be supplied initially.
This isn't literally the activation energy for the reaction, because atomisation energy and ionisation energy have quite specific meanings which don't apply here. But the actual activation energy will be affected by these two factors - the need to break the atom away from the lattice and its ionisation.
Note: Atomisation energy measures the energy needed to convert the solid metal into gaseous scattered atoms, and that is clearly not what is happening here.
Similarly ionisation energy measures the energy needed to remove an electron from the atoms to produce positive ions - again in the gas state. That's not the case here either.
Both of these things become easier as you go down the group, and so the reactions get faster.
So although lithium releases most heat during the reaction, it does it relatively slowly - it isn't all released in one short, sharp burst. With caesium, on the other hand, although it doesn't release quite as much heat overall, it does it extremely quickly - and you get an explosion.
Summarising the reason for the increase in reactivity as you go down the Group
The reactions become easier as the energy needed to form positive ions falls. This is in part due to a decrease in ionisation energy as you go down the Group, and in part to a fall in atomisation energy reflecting weaker metallic bonds as you go from lithium to caesium. This leads to lower activation energies, and therefore faster reactions.
Note: If you are a UK A level student, you will almost certainly find that your examiners will only expect you to explain this in terms of the fall in ionisation energy as you go down the Group. In other words, they simplify things by overlooking the contribution from atomisation energy. Stick with what your examiners expect - don't make life difficult for yourself! I'm trying to be as rigorous as I can because a sizeable part of my audience is working in systems outside the UK or beyond A level.
© Jim Clark 2005 (last modified January 2024)