Second Law of ThermodynamicsIn 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
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".
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:
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.)
Note that multipling the coefficient (which is in moles) by the unit for molar enthalpy (J/mol·K) leaves the unit of J/K Step 2: sum the product molecules, and the reactant molecules
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.