In chemical reactions entropy can be considered to be a measure of the number of ways which molecules and their quanta of energy can be arranged and is measured in J K-1 mol-1.
For a process to occur spontaneously the total entropy must always increase (This is the second law of thermodynamics). The entropy change of a reaction can be used to decide whether or not a reaction will occur spontaneously at a certain temperature. Bear in mind that this gives no indication of how quickly a reaction can occur only that it will; the reaction rate may be so low that the reaction isn’t noticeable/measurable.
In general solids have lower entropies than liquids which have lower entropies than gases. This can be explained very simply by looking at the molecular arrangements in each phase:
- In a solid the molecules are usually arranged in some form of lattice structure whereby movement of the molecules is severely restricted. This means that there are very few ways which the molecules can be arranged in the space and so the entropy is low.
- In a liquid the molecules are arranged randomly and are less restricted in their movements, this therefore means that they have higher entropies than solids as the molecules can be arranged more ways in the space.
- In gases the molecules are also randomly distributed in the space and are even more diffuse than in liquids with very little restriction to their movements in comparison. There are therefore more ways than in either solids or liquids for the molecules to be arranged in the space and so there entropies are generally the highest.
You can use the above rules to make a very basic/simplified decision on whether a reaction will occur spontaneously just by looking at the relative phases of the reactants and products, for example in the reaction below:
A(l) + B(l) → 2C(g)
Two liquids react to give a gas, gases have higher entropies than liquids and so by the above rules the entropy change of this reaction will be positive and so it will occur spontaneously.
Considering the surroundings
So far we have considered the entropy change of the system (ΔSsystem = ΣSproducts - ΣSreactants). What hasn’t been considered here however is that entropy is also a measure of the arrangement of quanta of energy not just the distribution of molecules. The energy changes associated with a process must also therefore be factored in.
An equation you may have come across is ΔSsurroundings = -ΔH/T. This equation allows us to consider the entropy change of the surroundings during a process rather than just that of the system. It is calculated by dividing the negative of the enthalpy change of the process by the temperature.
- This formula basically allows us to consider the distribution of the quanta of energy rather than just that of the molecules. It thus explains why processes occur which would seem impossible by just considering the differences between the starting and final molecular distributions. An example of this may be the freezing of water at -10oC. The idea of a liquid spontaneously turning into a solid would seem impossible by just considering the molecular distribution. It is moving from a less ordered to a more ordered state, however by considering the total entropy change of the process it can be rationalised.
The total entropy change of a reaction is calculated by ΔStotal = ΔSsystem - ΔSsurroundings. Using this equation you can ascertain whether a reaction will occur spontaneously at a given temperature.
An example of this is the freezing of water. This is a process, the temperature dependence of which we come across on a daily basis (water freezes at 0oC or 273K). This temperature dependence can be seen very clearly by looking at the total entropy change of the process (ΔStotal = ΔSsystem + ΔSsurroundings). The enthalpy change of water freezing is about ΔHo = -6010 J mol-1:
This means that ΔSsurroundings = -ΔH/T = - (-6010 J mol-1 / T) = +6010 J mol-1 / T
For water freezing ΔSsystem = -22.0 J K-1 mol-1.
Therefore ΔStotal = ΔSsystem + ΔSsurroundings = -22.0 J K-1 mol-1 + (+6010 J mol-1 / T)
= -22.0 J K-1 mol-1 + 6010 J mol-1 / T).
At 263K (-10oC) ΔStotal = -22.0 J K-1 mol-1 + (6010 J mol-1 / 263K)
= -22.0 J K-1 mol-1 + (6010 J mol-1 / 283K) = -22.0 J K-1 mol-1 + 22.9 J K-1 mol-1 = +0.9 J mol-1 K-1
The entropy change of the process is positive and so the process will occur spontaneously at this temperature (water does freeze at -10oC).
If we look at whether water will freeze at 283K (+10oC) however, we get a different story:
ΔStotal = ΔSsystem + ΔSsurroundings = -22.0 J K-1 mol-1 + (+6010 J mol-1 / T)
ΔStotal = -22.0 J K-1 mol-1 + (6010 J mol-1 / 283K)
= -22.0 J K-1 mol-1 + (6010 J mol-1 / 283K) = -22.0 J K-1 mol-1 + 21.2 J K-1 mol-1 = -0.8 J K-1 mol-1
The entropy change of the process of water freezing at 283K is negative, therefore this process will not occur spontaneously (water doesn’t freeze at +10oC).
As you can see, using entropy calculations we can rationalise the temperature at which processes happen, the freezing of water in this case. These calculations can also be used to find the minimum/threshold temperature at which a process will occur spontaneously (the temperature whereby ΔStotal = 0). This is the temperature at which the phase change takes place, the point of equilibrium between the phases.
At higher level…
If you study chemistry to a higher level you will find that these calculations are simplified. There are also entropy changes associated with the difference in the temperature that the process takes place (-/+10oC) and that at which the phase change takes place (0oC). The diagrams used for mapping out the total change in entropy for the process, taking these extra changes into account, are similar to the Hess cycles you may have already come across.
All data for the calculations of the entropy changes of water freezing used from Chemical Ideas Third Edition (2008) Pearson Education Limited 2008 ISBN 978 0 435631 49 9