The First Law of Thermodynamics concluded that all systems have internal energy U. Common sense tells us this has to be so. This internal energy is the sum total of all the energies of all the activities that are in operation in the molecules that made up the system.

Now we are interested in the relationship between the molecules in a system. We are convinced that they would like to have their own space. They wanted to enjoy a certain degree of freedom. This is not difficult to accept. We human beings also like to have our own space. We like to be acknowledged as an individual of the group; with our own personalities.

Of course molecules are not human beings so to them being free is the only right they asked for. Many words have been used to express this tendency in a system. Words like randomness or chaos. The formal term is entropy, the energy (thermodynamics is strictly about energy in systems) that allow the system to have its "freedom". The symbol used is S. We proposed that this entropy is a state function and is a function of the heat energy in the system; that is ΔS α Δq.

Let us illustrate this with the phase transition (change in state at constant temperature) of a compound from a liquid phase to a vapour phase. We know that in the gas phase the molecules have more freedom than when it is in the liquid phase. So the entropy should increase. The heat of vapourisation is always positive (heat is added to the system).

If it is the other way round; the compound condenses, then heat must be given off by the system. The change of enthalpy ΔH will be negative and so entropy will be negative. The molecules now lose their freedom.

Trouton observed that if the heat of vapourisation was divided by the temperature (in Kelvin) the value is more or less 86 JK‾� mol‾�

The values for the non-polar argon and methane is because their is relatively more freedom for the molecules in the liquid phase, since there is very minimum forces of attraction. The value for water is much higher because the molecules in the liquid phase have less freedom being held together by hydrogen bonding.

This appears to be a good parameter to measure the "energy for freedom" at a constant temperature. So entropy is expressed as ΔS = Δq / T.

As I have never fail to state; thermodynamics is more that a science, it is a philosophy. We are trying to understand the laws governing the universe. So all the above comes from many years of pure wisdom from human experiences. We do not try to prove it. The only proof is by its results. So do not try to figure out how we come up with the concept of entropy or how we come to the expression. They are the product of the wisdom of the ages.

SECOND LAW of THERMODYNAMICS

The Second Law of Thermodynamics states that for a spontaneous process to occur in an isolated system, the entropy must increase. That is ΔS > 0. An isolated system is a system to and from which no energy or matter can flow. That is both closed and adiabatic.

J.W. Gibbs

This means that TΔS > Δq. To translate this philosophy to a science Josiah Willard Gibbs, the father of chemical thermodynamics, proposed a term G such that G = H − TS. At constant temperature and pressure, the conditions most chemistry are being conducted,

ΔG = ΔH − TΔS.

Since a spontaneous process would take place when 0 > Δq − TΔS, then Gibb's expression says that a chemical reaction at constant temperature and pressure could take place when ΔG < 0. In his honour G is now known as the Gibb's energy, or sometimes known as the Free Energy. You can think of it as the energy that measures whether the system can break free.

Note: If you believe that the universe itself is an isolated system then you can say that the entropy of the universe never decreases, but can only increase. What is the implication?

HOW DO WE MEASURE THE ENTROPY OF A COMPOUND at TEMPERATURE T?

Let us assume that the entropy of the compound at temperature 0 K is S(0), and the heat capacities at constant pressure are;

	solid	Liquid	Gas
Cp	Cp(s)	Cp(l)	Cp(g)

Then the entropy of the compound at any temperature T would be given by;

T(f)

Cp(s)

ΔHfus

T(b)

Cp(l)

ΔHvap

T

Cp(g)

S(T)

=

S(0)

+

∫

dT

+

+

∫

dT

+

+

∫

dT

0

T

Tfus

T(f)

T

Tvap

T(b)

T

					T	Cp(s)			ΔHfus		ΔHvap
	S(T)	=	S(0)	+	∫		dT	+		+
					0	T			Tfus		Tvap

We then conduct experiments to measure the change in heat capacities at constant pressure from T = 0 to T, and then plot (C_p/T) versus T, or C_p versus lnT. The area under the plot should give the value for the third term on the right hand side of the equation. (ΔH_fus, T_fus), and (ΔH_vap, T_vap) can be easily determined.

Tutorial 1

The standard entropy at 298 K and the variation of heat capacity with temperature (National Institute of Standard and Technology) for the following compounds are;

Compounds	n-Hexane	Propane	Propene	Neo-hexane
S� / J K‾� mol‾�	389	270	267	359
Temp. / K	Cp / J mol‾�
200 300 400 500 600 700 800	111 143 181 217 248 274 296	56.1 73.9 94.0 113 129 143 155	50.2 64.6 80.5 95.2 108 119 129	101 142 183 220 253 282 307

Compute S� of the compounds at 700 K.      Answer

The difficult segment of this experiment is the region close to T=0 K. It is not possible to work at such a low temperature. The lowest we can come to is about 10 K. Debye proposed that for this short interval the error is insignificant if we assumed ΔS = ⅓ C_p(low), where C_p(low) is the heat capacity at the lowest temperature we can measure. This is common referred to as the Debye extrapolation.

Tutorial 2

The molar heat capacity at constant pressure for N₂ at T = 10 K was determined as 5.76 JK‾� mol‾�, what is the entropy for N₂ at 10 K?    Answer

The only term left is S(0), the entropy at T = 0 K. At this temperature we believed that all movements in the molecules cease. The party is over because it is too cold for life to exist.

The Third Law of Thermodynamics states that at 0 K the entropy of a compound in their perfect crystalline state is zero. Sometimes when frozen there is still some molecules trapped in different alignment. Because of this the concept of entropy considers it as not perfectly uniform and so it is random.

An example will be carbon monoxide. There are two possible alignments for this compound. Here the center molecule is wrongly aligned (a defect in crystallisation) and so there should be some entropy in the solid.

It is important to note that while we normally obtain tabulated ethalpy of formation, Δ_fH, and free energy of formation Δ_fG, we seldom tabulate entropy of formation. Instead the data are for the entropy of elements and compounds. We are expected to compute for the entropy of formation.

Tutorial 3

For methane gas at 298 K; Δ_fH = −74.81 kJ mol‾� and Δ_fG = −50.72 kJ mol‾�. What is the entropy for the formation of methane gas at 298 K?

Given the entropy at 298 K for graphite = 5.74 J mol‾�, hydrogen gas = 130.68 J mol‾�, and methane = 186.26 J mol‾�; what is the entropy of formation of methane gas at 298 K?      Answer

Reference

Thermodynamics data of compounds: National Institute of Standard and Technology