16.4 Third Law of Thermodynamics
A crystal is a substance that is made up of small unit cells, which are arranged in a highly ordered manner. This picture on the right shows the regular arrangement of particles in a crystal. These particles are allowed to vibrate in their place only. Since entropy is the measure of randomness, you can say a crystal will have very low entropy.
As we lower the temperature, the movement of particles gets further restricted. At absolute zero 0 K, we say that particles’ motion will virtually stop, and hence there will be no randomness at all in the crystal at that temperature.
As we discussed earlier, the relation between a system’s entropy and the number of possible microstates given by Boltzmann’s equation is:
S = k ln W
Here k is the Boltzmann constant, 1.38 × 10−23 J/K.
At 0 K, since there is no randomness left, we can safely assume that there will be only one microstate possible for the system, or W = 1, putting this in above equation we will get
S = 1.38 × 10−23 ln 1
Since ln 1 = 1, we will get S = 1.38 ×10−23 × 1 = 0
This condition of temperature can be summarized in the form of third law of Thermodynamics.
According to the third law of thermodynamics, the entropy of a perfect crystal at 0 K is zero. With the increase in temperature, particles’ vibrational motion will increase, and hence the entropy will also increase.
Unit 5: Free Energy Change for Coupled Reactions
Free energy is a state function. That means it only depends on the object’s current state and not on the path taken to achieve that state. Recall your knowledge from the earlier thermochemistry chapter, we mentioned that enthalpy is a state function and we can use Hess’s law to find the enthalpy of the reaction with the help of other reactions. A similar concept can be applied to the calculation of free energy as well.
Question 13: Find the free energy change ∆G0 for the reaction CDiamond CGraphite
Given: Cgraphite (s) + O2 (g) 
CO2 (g) ∆G0 = -394 kj/mol
CDiamond (s) + O2 (g) CO2 (g) ∆G0 = –397 kj/mol
We will flip the first reaction and then add these two reactions to get the desired reaction.
CO2 (g) Cgraphite (s) + O2 (g) ∆G0 = 394 kJ/mol
CDiamond (s) + O2 (g) CO2 (g) ∆G0 = –397 kj/mol
CDiamond CGraphite ∆G0 = – 3 kj/mol
Question 14:
Given
2C2H2 (g) + 5O2 (g) 4CO2 (g) + 2H2O (l) ∆G0 =-2470.04 kJ
2C2H6 (g) + 7O2 (g) 4CO2 (g) + 6H2O (l) ∆G0 =-2936.04 kJ
2H2 (g) + O2 (g) 2H2O (l) ∆G0 =-474.2 kJ
Based on the reactions above find the ∆G0 for the following reaction.
C2H2 (g) + 2H2 (g) C2H6(g)
Answer: We need C2H2 on the reactant side, so use the first reaction as such. Since we need C2H6 on the product side we can flip reaction 2. Reactions 1 and 2 both have 2 times as many C2H2 andC2H6 as needed in the final reaction, so we can multiply reaction 3 also with 2, so all the reactants and products are now double the desired quantity. Adding these reactions now can give us an overall reaction with twice the amount of reactants and products.
2C2H2 (g) + 5O2 (g) 4CO2 (g) + 2H2O (l) ∆G0 =-2470.04 kJ
4CO2 (g) + 6H2O (l) 2C2H6 (g) + 7O2 (g) ∆G0 = 2936.04 kJ
4H2 (g) + 2 O2 (g) 4H2O (l) ∆G0 =-474.2 x 2 kJ
2C2H2 (g) + 4H2 (g) 2C2H6(g) ∆G0 =-482.4 kJ
We can divide this whole reaction by 2 to get the desired reaction.
C2H2 (g) + 2H2 (g) C2H6(g) ∆G0 =-241.2 kJ
∆G0 value can help us predict if a reaction can occur spontaneously or not. However, how fast or slow the reaction will occur only depends on the kinetics of the reaction. It can be the browning of the apple or the conversion of diamond to graphite, thermodynamics only help us predict the spontaneity of the reactions.