If you flip two coins there are three possible outcomes. Both coins could be heads, both could be tails or you could have one of each. After several trials, if you examine the data you will notice that the probability of getting one of each is about twice that of getting two heads or two tails. We could express the ratio of the probabilities as 1 HH: 2 HT: 1TT.

Why is the probability of getting one of each twice that of getting both heads or both tails? If we look at the number of ways the result can be two heads, there is only one way that can happen. Getting one head and one tail in a flip has two ways of occurring, HT or TH. We can think of HT and TH as microstates of the one macrostate, n = 1.

Any occurrence will be more likely to occur if there are multiple ways of achieving it. Changes that occur tend to have a higher probability of occurring if there are multiple ways achieving that outcome.

Entropy (S)

The scientific term for the number of ways for something to occur is entropy. As we saw with the rock-paper-scissors activity, the probability of everyone ending up with the same amount of energy was low, whereas the probability of other arrangements was greater because there was multiple ways for it to occur. Flipping 10 coins would produce the 5 Heads outcome with greater frequency because there are more ways for it to occur. We would expect a very low frequency of getting 10 heads or 10 tails simply because there is only one way for each to occur. As with flipping coins or rock-paper-scissors, spontaneous changes in nature statistically go towards a greater number of ways (W_f > W_i), or a greater amount of entropy.

The previous unit took a kinetic or particle view of the flow of energy. Our focus in this unit will be on probability. Changes in nature obey the law of conservation of mass and energy. Every process must obey these rules. However for every process (i.e. combustion of gasoline and O₂ into CO₂ and H₂O) there is a reverse process (i.e. reacting CO₂ and H₂O to form gasoline and O₂), which also conserves mass and energy. While both processes are possible, only one is probable. The combustion reaction is much more probable because the process results in an increase in total amount of entropy.

 In more complex situations, we need to consider both system, surroundings, particles and energy. In this situation the total entropy change would be the sum of the entropy changes for each part: DS_total = DS₁ + DS₂ + DS₃ + … When DS_total >0, the process is so likely that it borders on certainty. When DS_total < 0, the process is so unlikely that the only the reverse process can only occur. When DS_total = 0, both directions are equally likely and the system is in equilibrium.

Thermal Processes

Let’s consider DS for thermal processes. We define the quantity DS_therm as a measure of the number of ways in which thermal energy can be distributed in the object. When a hot object is placed in contact with a cold one, there is a net flow of energy (Q) from the hotter to the colder object. Since the hot object loses energy Q is (-). For the cold object Q is (+). If the system is insulated |Q_hot| = |Q_cold|.

Since the hot object loses energy it has less remaining energy to be distributed, so DS_hot must be (-). The colder object gains energy and now has more energy to be distributed, so DS_cold must be (+). The total change in entropy for the system is:

DS_hot + DS_cold = DS_total

                                                            (-)         (+)

Since the process is spontaneous, DS_total must be (+). Because both objects have the same magnitude Q, there must be another factor that affects entropy. This factor is the Kelvin temperature T. We can now define thermal entropy DS_th = Q/T. Therefore:

                                             DS_hot  =    Q_hot        DS_cold  =  Q_cold  

                                                                T_hot                          T_cold

Since T_hot > T_cold, and | Q_hot| =| Q_cold|, |DS_hot| must be less than |DS_cold| and

                                                  DS_total =   DS_hot + DS_cold = (+)

                                                                        (-)         (+)

• Let Q_hot = - 100 J
• Let Q_cold = + 100 J
• Let T_cold = 10 K
• Let T_hot = 100 K

DS_hot  =    Q_hot    = -100 J     = -1 J/K                     

                  T_hot         100 K     

DS_cold  =  Q_cold  = +100 J = +10 J/K

                 T_cold          10 K

DS_total= DS_hot + DS_cold = -1 + 10 = +9 J/K

The process of moving energy from hot to cold is spontaneous and has a positive DS_total. The reverse process, moving 100 J of energy from cold to hot would be non-spontaneous and have a negative DS_total. Why?

Structural / Positional Change

We have learned that if the temperature of the system and the surroundings are different, there is a higher probability that energy will flow from a higher to a lower temperatureresulting in a positive DS_therm. Because energy can also be stored in the interaction account, we need to consider changes in  the arrangement and interaction of particles within the system.

Consider two containers connected by a tube that has a spigot on it that is closed. One container is filled with gas particles and the other container is empty. When we open the spigot, we know that some of the gas particles will move into the other container. Why does this occur spontaneously?

The gas particles move spontaneously to occupy both containers because the number of ways they can arrange themselves throughout two containers is more than in just one container. Below are just three examples of how the particles can be arranged differently.

Structural/Positional and Thermal Change

In a previous unit we discussed how energy is allocated during changes in state. In the icy hot lab, we noticed plateaus in the temperature as the changes in state were occuring. We had assumed that the added energy "decided" at the melting point to switch from entering the thermal account to entering the interaction account. Does the added energy really know where it is supposed to go at the melting point? In this unit we examined changes in state in more detail with the melting lauric acid lab and the icy cold lab. Our observations from both experiments led us to the conclusion that any energy that is added to the system enters the thermal account and then is transferred to the interaction account.  In fact, in the icy cold experiment and for many endothermic processes, energy doesn't need to be added from outside the system in order for the process to occur. Adding the salt to the ice resulted in melting without the addition of energy from the surroundings. The decrease in temperature was the result of energy being transferred from the thermal account. 

This still leaves us the question of why the energy moves from the thermal to the interaction account.

When changes in state or chemical changes take place, the overall DS includes the DS_st and the DS_th.

DS_overall = DS_st + DS_th

DS_total =   DS_st + DS_th = (-)

                  (+)         (-)

In the icy cold experiment, we made melting favorable (overall DS positive and favorable) by adding the salt. As the salt dissolves, it breaks into ions and has many more ways of arranging itself and interacting with the water molecules.This increases the DS_st  enough to make the overall DS favorable

Without the addition of salt the ice would melt at the mormal melting temperature because the increase in T decreases the magnitude of Q/T, making DS_th a smaller negative. The result is a DS_overall that is positive and favorable.

DS_total =   DS_st + DS_th = (+)

                  (+)      (-)

The same reasoning used above applies to any phase change. For melting or boiling, DS_th is negative and DS_st is positive. For condensation and freezing, energy moves from the Ei account to the Eth account making Q, Q/T and DS_th positive. The DS_st is negative because the number of ways the particles can arrange themselves decreases.
 
.Why does the energy move between Eth and Ei only at the melting point? At that temperature, the change in state produces a large positive DS_st that contributes to an overall positive DS, which makes the transfer of energy from Eth to Ei favorable.  Addition of salt to the water makes for an even larger DSst, which allows for a larger DSth and a lower melting point.
 

Gibbs Free Energy

We can determine whether or not a reaction or change will take place spontaneously. In order to do this chemists have combined some of the terms we’ve learned about in this and earlier units into one equation that can be used to determine whether or not a reaction will proceed spontaneously.

In this equation,

• DG is Gibbs Free Energy (not the weatherman’s!) and is related to overall DS
• DH is the enthalpy change in a reaction (related to the DSth),
• T is the Kelvin temperature 
• DS is the change in entropy(DSst). 

If the calculated DG is negative, the reaction will proceed spontaneously. If DG is positive, the reverse reaction is spontaneous.

Lets look at an example.

Boiling Water:

 DH = +44 kJ/mol

DS = +0.1188 kJ/molK

Will water spontaneously boil at –10 C (263K)? at 150 C (423K)?

At –10 C:         DG = DH - TDS = +44 kJ/mol – 263K(0.1188kJ/molK) = +12.75 kJ/mol 

Boiling at –10 C is NOT spontaneous.

At 150 C         DG = DH - TDS = +44 kJ/mol – 423K(0.1188kJ/molK) = -6.25 kJ/mol   

Boiling at 150 C IS spontaneous.