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:
  1. 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.
  2. 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.
  3. 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.
  4. 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.

Useful books for revision