Entropy – a Further Look

Second Law of Thermodynamics

In any change, the entropy of the universe must increase.

We can consider the universe as being made of two parts:

Everything in the universe is either a part of the system or surroundings, so

Suniverse = Ssystem + Ssurroundings

Sokatisewin dam, Churchill River, Sandy Bay, Saskatchewan
Water falling over a power dam converts gravitational potential energy into kinetic energy to turn a turbine.  In turn this kinetic energy is converted into electrical energy.   However, some energy is lost as heat is dispersed into the universe – the entropy of the universe increases.  It must do so, or the water would not flow over the dam and do useful work.

In any change the entropy of the universe increases, so Suniverse > 0.   It should be pretty obvious that the entropy of the universe could increase if either (or both) of the entropies of the system or surroundings increased enough.  Individually it is possible for either Ssystem or Ssurroundings to be increasing or decreasing, but together the entropy of the universe will increase, if a change is taking place.

Water does spontaneously fall downwards, and when it does the Suniverse must increase.  As the water falls, its potential energy changes into kinetic energy.  When the water hits the ground, this kinetic energy will cause some heat – not a lot, of course, but it will warm up the surroundings a bit.  This heat energy is dispersed out into all the molecules of the surroundings, so that the Ssurroundings increases.  The net result is that Suniverse also increases, so the change takes place and water falls spontaneously down hill.  Even the idea that systems tend towards minimum energy is really just another consequence of the second law of thermodynamics.

The second law of thermodynamics tells us what happens in a spontaneous change, but like all scientific laws, doesn't explain why it does so.  The explanation for why there must be an increase in entropy came from Ludwig Boltzmann.  He showed that entropy is really the consequence of probability.   If you have molecules moving about at random, then it is more likely that they will become mixed than stay separate.  This is why you will sometimes hear entropy described as "randomness" or "disorder".  As entropy increases, the distribution of energies that can exist in molecules become more "mixed up".

A messy room is NOT an example of entropy.
A messy room is not a good example of entropy.  While it may appear to spontaneously get more disordered, it isn't because of an increase in entropy.  The concept of disorder and entropy only applies to small objects such as molecules.  It is really important to keep in mind that we're talking about the disorder of energy in the microscopic things we call molecules, and not anything larger.

So where does the energy get dispersed or spread-out into?  It goes into the atoms and molecules of which the substance is made.  Molecules have many different ways of dispersing the energy into all the different kinds of motion that they can do – rotation, translation, and vibration.  An increase in entropy is an increase in the number of ways in which heat energy is distributed amongst molecules.

Some important things to remember about entropy are:

Sometimes it is possible to just look at a change and identify immediately if entropy is increasing.  For example:

However, for other reactions it is not at all obvious.  For example, consider the following reaction, the combustion of methane

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

Here everything is gas, and we have mixtures of both elements and compounds.   While we might guess that the entropy of the two product compounds is greater than the one reactant compound and one reactant element, there is no sound reason to do this.   However, there is a way to calculate the entropy change in any reaction, at least at standard conditions of 25 oC and 1 bar (atmospheric pressure).  The standard molar entropies, So, of many compounds and elements have been recorded in tables of thermodynamic data.  These can be combined using the following relationship to find the entropy change for any reaction under standard conditions:

calcs_01.gif (455 bytes)

In other words, sum up all the molar entropies of all the product molecules and subtract from this the sum of all the molar entropies of all the reactant molecules (this is the standard definition for change: final state - initial state).

Here's the process for the combustion of methane.

Step 1: find all the values of So (product molecules are displayed against a purple background, and reactant molecules red.)

Find all the values of So for each molecule in the reaction 

Multiply by the reaction coefficient

CH4 (g) 186.25 1(186.25 J/mol·K)  =  186.25 J/K
2O2 (g) 205.152 2(206.152 J/mol·K) =  412.304 J/K
CO2 (g) 213.785 1(213.785 J/mol·K) =  213.785 J/K
2H2O (g) 188.835 2(188.835 J/mol·K) =  377.670 J/K

Note that multipling the coefficient (which is in moles) by the unit for molar enthalpy (J/mol·K) leaves the unit of J/K

calcs_02.gif (380 bytes)

Step 2: sum the product molecules, and the reactant molecules

Products Reactants
213.785 186.25
377.670 412.304
======
591.455 J/K
======
598.55  J/K

Step 3: subtract products minus reactants (careful: make sure it is products minus reactants).  Round off to the correct number of significant digits.

Ssystem = 591.455 J/K - 598.55 J/K = -7.10 J/K

There is actually a slight decrease in entropy for the system during this process.

Remember that what you have calculated is at the standard state.  The entropy will be different under different pressure and temperature conditions.

What is the standard entropy change for the following reaction:

CaCO3 (s) CaO (s) + CO2 (g)

S =  (in units of kJ/K)