Entropy and the Second Law of Thermodynamics

Entropy of the Surroundings

Why does water freeze at temperatures below 0^oC?

Water has a greater entropy than ice and so entropy favours melting.

But ice has a lower energy than water and so energy favours freezing.

It is possible to predict what will happen by taking into account the entropy of the surroundings, in addition to the energy of the system.

The following reaction shows the change in state from water to ice:

Freezing is an exothermic process; energy is lost from the water and dissipated to the surroundings.

Therefore, as the surroundings get hotter, they are gaining more energy and thus the entropy of the surroundings is increasing.

The amount by which the entropy of the surroundings has increased can be determined using the following principle: the entropy of the surroundings increases by an amount equal to the heat energy they gain divided by the temperature at which it happens, therefore:

This relationship allows us to make a prediction about the entropy of the surroundings of a chemical process, whatever they are (even the whole universe!), using the measurements we can make on the chemical system.

The total entropy change can then be used to predict whether a reaction is feasible or not at a given temperature.

The total entropy is equal to: 

Using this equation for the freezing of water at -10^oC for example: 

The overall entropy increases, and so the reaction is feasible (in thermodynamic terms, the reaction would be described as being spontaneous; however, this does not in any way describe how fast the reaction occurs).

The feasibility of the reaction can then be determined at 10^oC:

At the higher temperature, the reaction is not feasible as the entropy of the reaction is decreasing. However, ice will melt at this temperature, as the entropy increases :

Both the processes of ice melting at 10^oC and water freezing at -10^oC are spontaneous reactions; they can occur. Both these reactions occur as they result in an increase in the total entropy.
S_total is positive for a spontaneous reaction

This is part of the Second Law of Thermodynamics and it is very important as it allows us to predict whether something can happen or not.

Even though a reaction can be spontaneous, it is not certainty that the reaction will occur, as the activation energy may be too high for the reaction.

The second law of thermodynamics can be used to explain the following:

The reaction will occur, as in an exothermic reaction H is negative, and if the entropy increases, then S is positive, so:Total entropy change is positive, so reaction is feasible.

The reaction can never occur, as H is positive and S is negative:The total entropy change is negative and so the reaction cannot occur.

The reaction can occur depending on the conditions:As the temperature increases, S_surr increases (becomes less negative), therefore, the higher the temperature, the more likely it is for the reaction to occur. This ties in with La Chatelier’s principle- an increase in the temperature of an endothermic reaction pushes the position of equilibrium to the product side.

The reaction can occur depending on the conditions:As the temperature decreases, S_surr increases, therefore, the lower the temperature, the more likely it is for the reaction to occur.

Summary

When changes occur in a chemical reaction, they are nearly always accompanied by changes in the surroundings.

To predict whether or not a change will take place, we need to take account of the entropy changes in the system and its surroundings. 

For a change to be spontaneous, the total entropy must increase.

The Second Law of Thermodynamics states that for a spontaneous reaction S_total > 0.

Using the Second Law of Thermodynamics- Saltwater freezing

Have you ever wondered why salt water freezes at a lower temperature than pure water? Why icebergs are present in sea water and why putting salt on the road in winter prevents ice from forming?

This is all related to the process:

When seawater freezes, pure ice is produced; the energy released will be the same as when molecules from pure water fit together to form a lattice in an ice crystal.

H will still be -6.01 kJ mol^-1 when seawater freezes.

However, S for the overall system is different.

The entropy of the salt solution is greater than the entropy of pure water; the Na⁺ and Cl^- ions are spread out in the solution.

When salt water freezes, the S is more negative, due to the fact that the salt water is more disordered, and so it becomes more ordered when the water frezzes.

Therefore, the S_sys for salt water freezing is less than -22.0 J K^-1.

If the S_sys for salt water freezing is lower, then the S_surr has to be more positive than +22.0 kJ mol^-1 in order for the reaction to be spontaneous.

From the following equation: 

S_surr can only be greater than +22.0 kJ mol^-1 if we divide -H by a smaller value of T, so the freezing of salt water takes place at a temperature lower than 0^oC.

Why limestone doesn’t dissolve

At high temperatures, calcium carbonate can thermally decompose to form calcium oxide and carbon dioxide:

At normal temperatures on Earth, this process does not occur; this, once again, is related to the Second Law of Thermodynamics.

The following shows the entropy change at 25^oC (298K):

The total entropy change for this reaction is negative; therefore it is not spontaneous.

At 1273K, the entropy change is as follows:

At the higher temperature, the S_surr is less negative and so the S_total becomes positive, meaning that the reaction is feasible at the higher temperature.

Dissolving and Crystallisation

Why should salt dissolve to produce a solution (like sea water), but then crystallise from solution in the salt works when most of the water has evaporated?

The process of the salt dissolving is:

The entropy change is as follows:

At 298K, the entropy change is +26.1 J K^-1 mol^-1, so salt can dissolve to form a solution at 25^oC.

If the solution is more concentrated, then the S_sys value will change substantially:

There is a smaller volume of water for the ions to be dispersed in, thus there will be fewer possible arrangements of their positions.

The water molecules will hydrate the ions, and will therefore become more organised, rather than randomised in the liquid.

Therefore, the organised water molecules will make up a greater proportion of the water present.

Both of these result in the S_sys decreasing for the more concentrated solution.

Eventually the S_sys becomes less than 13.1 J K^-1 mol^-1, which leads to the S_total becoming negative, and so dissolving becomes unfavourable and crystals form from the solution.