16.1 Spontaneity and Second Law of Thermodynamics

A spontaneous process is a process that occurs naturally on its own without any additional aid. Let’s see some of the processes that take place around us on their own.  Water flowing down the mountain or a rock rolling down the hill are excellent examples of spontaneous processes. These reactions are spontaneous in one direction while nonspontaneous in another direction.  This rock does not go back to the top of the hill on its own; similarly, water does not climb back the mountain on its own without some additional aid.

Some of the processes are spontaneous in one direction or the other depending on the conditions around them.

An ice cube sitting outside the freezer on a hot sunny day will melt on its own. At the same time, water on the driveway will freeze on its own on a cold winter day. 

H₂O (s)                                  H₂O (l)

This reaction is spontaneous in the forward as well as backward direction, depending on the temperature.  So, the question is, why temperature change is making this process spontaneous in the forward or backward direction?

After looking into this question deeply and studying several reactions, scientists realized that it is entropy that decides the spontaneity of the reaction.

In simple words, entropy is the measure of randomness or disorder, often shown with the letter S. The total entropy of the universe is continuously increasing. The second law of thermodynamics states that a process is spontaneous if it increases the entropy of the universe.

Recall your memory from the earlier chapter on thermochemistry.  The universe is made up of the system and the surrounding. Therefore, as per the second law of thermodynamics, we can say that the sum of the change in entropy of the system and the surrounding will be positive for a spontaneous reaction.

Look at these three containers in the picture below. The first container has solid particles arranged regularly in a definite pattern.  If you look at the middle container, it has liquid particles, more disorderly than the first one. And the last container has particles in the gas phase.

Source: www.openstax.org/

As we increase the temperature of the solid state, the kinetic energy of particles increases, and they start vibrating faster. Once they have sufficient energy to overcome some of the intermolecular forces between particles, they get more disorderly, and the solid-state changes to a liquid.  Further increase in temperature will increase the energy of these particles some more and they will overcome remaining intermolecular forces and become independent of each other and change to the gas phase. Here, in the picture above we can see gas particles are randomly distributed all over the container.  Since the disorder is maximum in gas phase particles, we can say entropy will be maximum for the third container, and the first container with solid-phase particles will have the least entropy.

We can also understand the concept of entropy with the help of probability.  The higher the number of the arrangement of particles available, the higher the entropy will be.

If you look at this picture on the right.  There is just one combination possible when we have two separate compartments, one with two pink and one with two black molecules.  However, if we connect these two containers and allow particles to move, we can have several different arrangements possible for these molecules.  Few of those arrangements are shown here in the picture on the right.

Since a greater number of arrangements of molecules is possible in joined compartments, we can conclude that entropy will also be higher in joined compartments.  This type of entropy, which depends on the number of possible positions of particles, is known as positional entropy.

The same concept can be applied to a gas enclosed in a container.  If we allow this gas to expand, there are many more places to find gas particles, and thus entropy will increase with an increase in the volume of gas. Since an increase in entropy favors spontaneity, gases expand spontaneously and fill the entire container.

In general, any process that increases the number of microstates will increase the entropy of the system. A microstate is defined as a specific configuration of all the locations and energies of the atoms or molecules that make up a system. Look at the example below, we can see how a greater number of microstates can be associated with greater entropy.

In the picture below, the sixteen microstates associated with placing four different particles in two boxes are shown. The microstates are collected into five distributions—(a), (b), (c), (d), and (e)—based on the number of particles in each box.

Source: www.openstax.org/

As we can see, the most probable configuration is one of the six microstates associated with distribution (c) where the particles are evenly distributed between the boxes, that is, a configuration of two particles in each box. The probability of finding the system in this configuration is 1/16. Also, we can see in the picture above that the least probable distribution is shown in options a and e, where all 4 particles are in one box, and each has a probability of . Therefore, the combined probability of finding all four particles in one box, either the left box or right box is 1/16+ 1/16= 1/8. By this description, and combining it with the concept of entropy, we can say microstates in which all the particles are in a single box are the most ordered, thus possessing the least entropy (pictures a and e). And microstates in which the particles are more evenly distributed among the boxes are more disordered, possessing greater entropy, as shown in picture c.

Ludwig Boltzmann developed a mathematical relation between the entropy of a system (S) and the number of microstates (W) possible for the system. The relation between a system’s entropy and the number of possible microstates is:

S = k ln W

Here k is the Boltzmann constant, 1.38 ×× 10⁻²³ J/K.

Since we can calculate the change in entropy for a process by finding the difference between final entropy (S_f) and initial (S_i), we can say:

ΔS = S_f − S_i = k ln W_f – k ln W_i =

Here W_f is the final number of microstates and W_i is the initial number of microstates possible for the system.

Example: Find the change in entropy for the following.

In the picture above Initial number of microstates W_i = 6,

and Final number of microstates W_f = 1

Since ΔS =

Putting the value of Boltzmann constant, k = 1.38 ×× 10⁻²³ J/K, we will get

ΔS = 1.38 × 10⁻²³  = -2.47 × 10⁻²³ J/K

This change in entropy with a negative sign is consistent with the concept we saw above. As the number of microstates goes down, entropy decreases. We can see in the example above, the system was going from a higher number of microstates (6) to a lower number of microstates (1) possible, or from a more disordered state to a less disordered state, hence entropy decreased.

This concept of the number of microstates can also be applied to understand the entropy change with phase change in the example we discussed earlier. Solid state has particles in a fixed position, so the least number of microstates are available, hence the least entropy.  With the increase in kinetic energy and higher movement of particles possible in liquid state, we get more microstates possible, and entropy increases.  The gas phase has the highest kinetic energy and particles are allowed to move freely. So maximum number of microstates are possible. Hence particles have the highest entropy in gas phase.

Examples: Here are a few examples of reactions resulting in an increase in entropy as we move from the reactant to the product side.

1. Br₂ (s)                              Br₂ (l) 

Solid going to the liquid phase, particles will get more randomly distributed; hence entropy will increase.

2. NH₃ (g)                            N₂ (g)  + 3H₂ (g)

Looking at the mole ratio, we have 2 mols of NH₃ gas producing a total of 4 mols of gases (1 mol N₂ and 3 mol H₂).  Since the total number of particles is increasing, so will the entropy.

• NaCl (s)                               NaCl (aq)

In this process, we dissolve salt in water. In solution, particles get distributed all over the solution and can be found in more microstates than in the solid phase; hence entropy will increase.

• CO₂ (g) (1 atm)                                  CO₂ (g) (0.5 atm )

In this example, the gas’s pressure decreases from left to right; a decrease in pressure will allow gas particles to get more randomly distributed and, therefore, increase entropy.

•  NH₃ (g) (125 ⁰C)                             NH₃ (g) (175 ⁰C)

With the increase in temperature, particles’ kinetic energy will increase, and so will the disorder.  Hence entropy will increase.

Look at this youtube video to further understand the concept of entropy change.

To practice some problems on this concept, look at the video below.

Learning Check:

1. Which of the following processes will result in an increase in entropy?
2. N₂ (g) + 2H₂ (g)                    N₂H₄ (g)
3. N₂ (g)                       2N (g)
4. I₂ (s)                          I₂ (g) 
5. PCl₅ (s)                         PCl₃(s) + Cl₂ (g)           
6. Three moles of reactant yield one mole of product. Reaction a shows a decrease in entropy.  
7. One mole of N₂ yields two moles of elemental nitrogen, and an increase in the number of particles shows an increase in entropy.  
8. In reaction c, Solid Iodine sublimes and produce gaseous Iodine, which means more disorder, hence ∆S increases. 
9. In this reaction, 1 mole of solid PCl₅ yields one mole of solid PCl₃ and a mole of gaseous Cl₂. We are going from 1 mole of reactant to 2 moles of product and also a solid reactant is producing a gaseous product.  Both factors will contribute to an increase in entropy.

From the second law of thermodynamics, an increase in the entropy of the universe will always result in a spontaneous process. 

Based on this law, we can say, in a chemical reaction, if ∆S_universe is positive, the reaction will be spontaneous, and if ∆S_universe is negative, the reaction will be nonspontaneous under the given set of conditions.  If ∆S_universe is zero, then we can say the system will be at equilibrium.  We can summarize this concept in the table below.

Since   ∆S_universe = ∆S_system +∆S_surrounding

Breaking the universe in the system and the surrounding can further simplify this concept in this table.

So how do we decide the change in entropy of the system and surroundings?

Earlier in this unit, we saw how to predict the sign for ∆S_system. 

We can calculate the magnitude of the ∆S_system with thermodynamic reference data tables and the formula below.

∆S_system = ∑∆S(products) – ∑∆S(reactants)

You can use this link below for the thermodynamic data table.

https://www.drjez.com/uco/ChemTools/Standard%20Thermodynamic%20Values.pdf

Since the surrounding is not made up of reactants and products, above formula cannot be used to calculate ∆S_surrounding.  However, ∆S_surrounding is directly related to the enthalpy change of the system. If you recall your knowledge from thermochemistry, enthalpy, or the amount of energy absorbed or released by the system, is always equal to the amount of energy released or absorbed by the surrounding.  A reaction that is exothermic with respect to the system will be endothermic with respect to the surrounding, and a reaction that is endothermic with respect to the system will be exothermic with respect to the surrounding. However the magnitude of energy absorbed or released will be same for both the system and the surrounding.   Based on this fact, we can conclude that in an exothermic reaction, the energy released by the system will be absorbed by the surrounding.  As a result, the kinetic energy of the surrounding will increase.  And an increase in the kinetic energy of the surroundings will increase the entropy of the surroundings. Similarly, in an endothermic reaction, the kinetic energy of the surroundings will decrease, and as a result, the entropy of the surrounding will also decrease.

∆S_surrounding depends on the temperature and can be calculated by using the formula below.

∆S_surrounding =

Here, ∆H_(system) is the change in enthalpy of the system.  We put a negative sign to compensate for the fact that the change in enthalpy of the surrounding is always the opposite of the change in enthalpy of the system. Here, T is the temperature in Kelvin.

The magnitude of ∆H of the system can be calculated easily by using thermodynamic reference data tables and the formula below.

∆H_(system) = ∑∆H(products) – ∑∆H(reactants)

In this formula

∆S_universe = ∆S_system +∆S_surrounding

we can substitute ∆S_surrounding  with  

∆S_universe = ∆S_system  +

Based on this formula above, we can conclude that, once we know ∆S_system and ∆H_(system), we can calculate the ∆S_universe and predict the spontaneity of the reaction.

Here is an example of the conversion of liquid water to water vapors.

H₂O (l)                                   H₂O (g)

We will look at the signs of entropy change for both system and surroundings.

Let’s start our discussion with ∆S_system, which refers to the change in entropy of the system. In this scenario, we are observing the transformation of liquid water into water vapors. As particles in the gas phase exhibit a greater degree of randomness than those in the liquid phase, the entropy of the system will increase during this transformation. Additionally, this change in entropy can be linked to a change in volume. Specifically, when comparing the volume of 1 mole of liquid water to that of 1 mole of water vapors, the latter will occupy a larger volume. This increased volume corresponds to a higher degree of disorder, leading us to the conclusion that ∆S_system will be positive during this process.

To decide the sign for ∆S_surrounding, we will look at ∆H_system. The conversion of liquid water to water vapors is an endothermic reaction. Therefore, ∆H_(system) will be positive, and ∆H_surrounding will be negative, or surrounding will release energy.  As a result, the surrounding’s kinetic energy will decrease, and then the entropy of the surrounding will also decrease, or we can say ∆S_surrounding will be negative. 

Since ∆S_system is positive and ∆S_surrounding is negative,

and ∆S_universe = ∆S_system +∆S_surrounding ,

We can conclude that the reaction will be spontaneous when ∆S_system is greater in magnitude than∆S_surrounding.

Since ∆S_surrounding = as temperature T will increase ∆S_surrounding will get smaller in magnitude, and eventually, it will be smaller in magnitude than ∆S_system, and process will become spontaneous. These results are consistent with the observation that water spontaneously changes to vapors at a temperature equal to or greater than 100 ⁰C or 373.15 K.

Watch this video for a further understanding of this concept.

Question 1:  Find the change in entropy for the surrounding for the following reaction at 25 ⁰C.

CH₄ (g) + 2O₂ (g)                   CO₂ (g) + 2H₂O (l)  +  890.4 kJ

We know         ∆S_surrounding =

Since energy is given on the product side, it is an exothermic reaction with respect to the system.

Therefore, ∆H for the system is  =  – 890.4 kJ = – 890.4 kj x  = – 890400 J

Temperature T = 25 ⁰C

                       T = 25 + 273.15 = 298.15 K

∆S_surrounding = = 2986 J/ K

A positive sign indicates that the entropy of the surrounding will increase in this reaction.

Here is another example for calculations of entropy change of surrounding.

Learning Check:

Question 2: Find the entropy change for the system and the surrounding at 25 ⁰C for the following process.

Xe (g) + 2F₂ (g)                         XeF₄ (s)    ∆H = -251 kJ

Given:

Question 3: Calculate ∆S_universe and predict if the process above will be spontaneous at 25 ⁰C.

Solution:  

Xe (g) + 2F₂ (g)                         XeF₄ (s)    ∆H = -251 kJ

Given:   

 ∆H = -251 kJ or -251,000J,   T = 25 + 273.15 = 298.15 K

We know         ∆S_surrounding =

∆S_surrounding =

∆S_surrounding = 841.9 J/mol.K

∆S_system = ∑S_products – ∑S_reactants 

Given: ∆S _Xe = 169.7 J/K. mol, ∆S_F2 = 202.8 J/K. mol and ∆S _XeF4 = 146 J/K. mol

∆S_system = [∆S _XeF4 ] – [∆S _Xe + 2 ∆S _F2 ]

∆S_system = [146] – [169.7 + 2(202.8)] = -429.3 J/K. mol

3) ∆S_universe = ∆S_system +∆S_surrounding = 841.9 + -429.3 = 413 J/K. mol,

The reaction will be spontaneous because ∆S_universe is positive.